Identities

Q1: Multiple Choice Type:

i. If a is positive and \(a^2+\frac{1}{a^2}=18\); then the value of \(a-\frac{1}{a}\) is:

Step 1: Recall identity: \((a – \frac{1}{a})^2 = a^2 + \frac{1}{a^2} – 2\)
Step 2: Substitute given value:
\((a – \frac{1}{a})^2 = 18 – 2 = 16\)
Step 3: Solve for \(a – \frac{1}{a}\):
\(a – \frac{1}{a} = \sqrt{16} = 4\)
Answer: a. 4

ii. \(\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\) is equal to:

Step 1: Recall identities: \((x+y)(x-y) = x^2 – y^2\)
Step 2: Substitute:
\((x+y)(x-y)(x^2+y^2)(x^4+y^4) = (x^2 – y^2)(x^2 + y^2)(x^4 + y^4)\)
Step 3: Use \((x^2 – y^2)(x^2 + y^2) = x^4 – y^4\)
Then: \((x^4 – y^4)(x^4 + y^4) = x^8 – y^8\)
Answer: c. \(x^8 – y^8\)

iii. The value of \(102\times98\) is:

Step 1: Use identity: \((a+b)(a-b) = a^2 – b^2\)
Step 2: Let \(a = 100, b = 2\), then:
\(102 \times 98 = (100+2)(100-2) = 100^2 – 2^2\)
Step 3: Simplify:
\(102 \times 98 = 10000 – 4 = 9996\)
Answer: d. 9996

iv. \(\left(x+3\right)\left(x+3\right)-\left(x-2\right)\left(x-2\right)\) is equal to:

Step 1: Expand using \((a+b)^2 = a^2 + 2ab + b^2\)
\((x+3)^2 – (x-2)^2 = (x^2 + 6x + 9) – (x^2 – 4x + 4)\)
Step 2: Simplify:
\(x^2 + 6x + 9 – x^2 + 4x – 4 = 10x + 5\)
Answer: a. 10x + 5

v. If \(5a={30}^2-{25}^2\), the value of a is:

Step 1: Use \((a+b)(a-b) = a^2 – b^2\)
\(30^2 – 25^2 = (30+25)(30-25) = 55 \times 5 = 275\)
Step 2: Solve for a:
\(5a = 275 \Rightarrow a = \frac{275}{5} = 55\)
Answer: c. 55

vi. Statement 1: Cube of a binomial : \(\left(a-b\right)^3=a^3+3a^2b-3ab^2-b^3\)
Statement 2: \(\left(a-b\right)^2-\left(a+b\right)^2=4ab\).
Which of the following options is correct?

Step 1: Verify Statement 1:
\((a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3\) ❌ incorrect
Step 2: Verify Statement 2:
\((a-b)^2 – (a+b)^2 = (a^2 – 2ab + b^2) – (a^2 + 2ab + b^2) = -4ab\) ❌ incorrect
Answer: b. Both the statements are false.

vii. Assertion (A): Using appropriate identity, we get \(22.5\times21.5=484.75\)
Reason (R): The product : \(\left(x+y\right)\left(x-y\right)=x^2-y^2\)

Step 1: Let \(x = 22, y = 0.5\)
\((22.5 \times 21.5) = (22 + 0.5)(22 – 0.5) = 22^2 – 0.5^2 = 484 – 0.25 = 483.75\) ❌
Step 2: Correct identity usage: \((x+y)(x-y) = x^2 – y^2\)
\((22.5 + 21.5)/2 = 22\), but calculation shows \(22.5 \times 21.5 = 484.75\) ✅
Answer:d. A is false, but R is true.

viii. Assertion (A): If we add 9 with \(49x^2-42x\), the resultant expression will be a perfect square expression.
Reason (R): The product of the sum and difference of the same two terms = Difference of their squares.

Step 1: \(49x^2 – 42x + 9 = (7x – 3)^2\) ✅
Step 2: Reason statement refers to \((a+b)(a-b)=a^2-b^2\) ✅
Step 3: So, reason (R) is true but reason (R) is not the correct explanation of assertion (A).
Answer:b. Both A and R are correct, and R is not correct explanation for A.

ix. Assertion (A): If the volume of a cube is \(a^3+b^3+3ab(a+b)\), then the edge of the cube is \(a+b\)
Reason (R): \((\text{1st term} + \text{2nd term})^3 = (\text{1st term})^3 + 3(\text{1st term})^2(\text{2nd term}) + 3(\text{2nd term})^2(\text{1st term}) + (\text{2nd term})^3\)

Step 1: Identity: \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\) ✅
Step 2: Hence edge = \(a+b\) ✅
Answer:a. Both A and R are correct, and R is the correct explanation for A

x. Assertion (A): \(687 \times 687 – 313 \times 313 = 37400\)
Reason (R): The product of the sum and the difference of the same two terms = The square of their difference

Step 1: Identity: \((a+b)(a-b) = a^2 – b^2\) ✅
Step 2: Here, \(687 \times 687 – 313 \times 313 = (687+313)(687-313) = 1000 \times 374 = 374000\), not 37400 ❌
Answer:d. A is false, but R is true

Q2: Evaluate:

i. \((3x+\frac{1}{2})(2x+\frac{1}{3})\)

Step 1: Use distributive property: \((a+b)(c+d) = ac + ad + bc + bd\)
\((3x)(2x) + (3x)(\frac{1}{3}) + (\frac{1}{2})(2x) + (\frac{1}{2})(\frac{1}{3})\)
Step 2: Compute each term:
\(6x^2 + x + x + \frac{1}{6} = 6x^2 + 2x + \frac{1}{6}\)
Answer:\(6x^2 + 2x + \frac{1}{6}\)

ii. \((2a+0.5)(7a-0.3)\)

Step 1: Expand using distributive property:
\((2a)(7a) + (2a)(-0.3) + (0.5)(7a) + (0.5)(-0.3)\)
Step 2: Compute:
\(14a^2 – 0.6a + 3.5a – 0.15 = 14a^2 + 2.9a – 0.15\)
Answer:\(14a^2 + 2.9a – 0.15\)

iii. \((9-y)(7+y)\)

Step 1: Use \((a-b)(c+d) = ac + ad – bc – bd\)
\((9)(7) + (9)(y) – (y)(7) – (y)(y) = 63 + 9y – 7y – y^2\)
Step 2: Simplify:
\(63 + 2y – y^2 = -y^2 + 2y + 63\)
Answer:\(-y^2 + 2y + 63\)

iv. \((2-z)(15-z)\)

Step 1: Expand:
\(2 \cdot 15 – 2z – 15z + z^2 = 30 – 17z + z^2\)
Answer:\(z^2 – 17z + 30\)

v. \((a^2+5)(a^2-3)\)

Step 1: Use identity \((x+y)(x-y) = x^2 – y^2\) if applicable. Here \(a^2+5\) and \(a^2-3\) are not in same form, so expand directly:
\((a^2)(a^2) + (a^2)(-3) + 5(a^2) + 5(-3) = a^4 – 3a^2 + 5a^2 – 15\)
Step 2: Simplify:
\(a^4 + 2a^2 – 15\)
Answer:\(a^4 + 2a^2 – 15\)

vi. \((4-ab)(8+ab)\)

Step 1: Use identity \((x-y)(x+y) = x^2 – y^2\) with \(x=4, y=ab-4\)?
Actually expand directly:
\((4)(8) + (4)(ab) – (ab)(8) – (ab)(ab) = 32 + 4ab – 8ab – a^2b^2\)
Step 2: Simplify:
\(32 – 4ab – a^2b^2\)
Answer:\(32 – 4ab – a^2 b^2\)

vii. \((5xy-7)(7xy+9)\)

Step 1: Expand using distributive property:
\((5xy)(7xy) + (5xy)(9) + (-7)(7xy) + (-7)(9)\)
Step 2: Compute each term:
\(35x^2y^2 + 45xy – 49xy – 63\)
Step 3: Simplify:
\(35x^2y^2 – 4xy – 63\)
Answer:\(35x^2 y^2 – 4xy – 63\)

viii. \((3a^2-4b^2)(8a^2-3b^2)\)

Step 1: Expand using distributive property:
\((3a^2)(8a^2) + (3a^2)(-3b^2) + (-4b^2)(8a^2) + (-4b^2)(-3b^2)\)
Step 2: Compute each term:
\(24a^4 – 9a^2b^2 – 32a^2b^2 + 12b^4\)
Step 3: Simplify:
\(24a^4 – 41a^2b^2 + 12b^4\)
Answer:\(24a^4 – 41a^2 b^2 + 12b^4\)

Q3: Find the square of:

i. \((3x + \frac{2}{y})^2\)

Step 1: Use identity \((a+b)^2 = a^2 + 2ab + b^2\)
Step 2: Substitute \(a = 3x, b = \frac{2}{y}\)
\((3x)^2 + 2(3x)(\frac{2}{y}) + (\frac{2}{y})^2\)
Step 3: Compute each term:
\(9x^2 + \frac{12x}{y} + \frac{4}{y^2}\)
Answer:\(9x^2 + \frac{12x}{y} + \frac{4}{y^2}\)

ii. \((\frac{5a}{6b} – \frac{6b}{5a})^2\)

Step 1: Use identity \((a-b)^2 = a^2 – 2ab + b^2\)
Step 2: Substitute \(a = \frac{5a}{6b}, b = \frac{6b}{5a}\)
\((\frac{5a}{6b})^2 – 2(\frac{5a}{6b})(\frac{6b}{5a}) + (\frac{6b}{5a})^2\)
Step 3: Compute each term:
\(\frac{25a^2}{36b^2} – 2(1) + \frac{36b^2}{25a^2} = \frac{25a^2}{36b^2} + \frac{36b^2}{25a^2} – 2\)
Answer:\(\frac{25a^2}{36b^2} + \frac{36b^2}{25a^2} – 2\)

iii. \((2m^2 – \frac{2}{3}n^2)^2\)

Step 1: \((a-b)^2 = a^2 – 2ab + b^2\)
Step 2: \(a = 2m^2, b = \frac{2}{3}n^2\)
\((2m^2)^2 – 2(2m^2)(\frac{2}{3}n^2) + (\frac{2}{3}n^2)^2\)
Step 3: Compute:
\(4m^4 – \frac{8}{3}m^2n^2 + \frac{4}{9}n^4\)
Answer:\(4m^4 – \frac{8}{3} m^2n^2 + \frac{4}{9} n^4\)

iv. \((5x + \frac{1}{5x})^2\)

Step 1: \((a+b)^2 = a^2 + 2ab + b^2\)
Step 2: \(a=5x, b=\frac{1}{5x}\)
\((5x)^2 + 2(5x)(\frac{1}{5x}) + (\frac{1}{5x})^2 = 25x^2 + 2 + \frac{1}{25x^2}\)
Answer:\(25x^2 + 2 + \frac{1}{25x^2}\)

v. \((8x + \frac{3}{2}y)^2\)

Step 1: \((a+b)^2 = a^2 + 2ab + b^2\)
\((8x)^2 + 2(8x)(\frac{3}{2}y) + (\frac{3}{2}y)^2 = 64x^2 + 24xy + \frac{9}{4}y^2\)
Answer:\(64x^2 + 24xy + \frac{9}{4} y^2\)

vi. \(607^2\)

Step 1: Use \((x+y)^2 = x^2 + 2xy + y^2\), take \(x=600, y=7\)
\((600+7)^2 = 600^2 + 2(600)(7) + 7^2 = 360000 + 8400 + 49 = 368449\)
Answer:368449

vii. \(391^2\)

Step 1: \(x=400, y=9, 391 = 400-9\), use \((x-y)^2 = x^2 – 2xy + y^2\)
\(400^2 – 2(400)(9) + 9^2 = 160000 – 7200 + 81 = 152881\)
Answer:152881

viii. \(9.7^2\)

Step 1: \(x=10, y=0.3, 9.7 = 10-0.3\)
\((10-0.3)^2 = 10^2 – 2(10)(0.3) + 0.3^2 = 100 – 6 + 0.09 = 94.09\)
Answer:94.09

Q4: If \(a+\frac{1}{a}=2\), find:

i. \(a^2+\frac{1}{a^2}\)

Step 1: Use identity: \((a+\frac{1}{a})^2 = a^2 + 2 + \frac{1}{a^2}\)
Step 2: Substitute \(a+\frac{1}{a} = 2\)
\((2)^2 = a^2 + 2 + \frac{1}{a^2}\)
Step 3: Simplify:
\(4 = a^2 + \frac{1}{a^2} + 2\)
\(a^2 + \frac{1}{a^2} = 4 – 2 = 2\)
Answer:2

ii. \(a^4+\frac{1}{a^4}\)

Step 1: Use identity: \((a^2+\frac{1}{a^2})^2 = a^4 + 2 + \frac{1}{a^4}\)
Step 2: Substitute \(a^2+\frac{1}{a^2} = 2\)
\((2)^2 = a^4 + 2 + \frac{1}{a^4}\)
Step 3: Simplify:
\(4 = a^4 + \frac{1}{a^4} + 2\)
\(a^4 + \frac{1}{a^4} = 4 – 2 = 2\)
Answer:2

Q5: If \(m-\frac{1}{m}=5\), find:

i. \(m^2+\frac{1}{m^2}\)

Step 1: Use identity: \((m-\frac{1}{m})^2 = m^2 – 2 + \frac{1}{m^2}\)
Step 2: Substitute \(m-\frac{1}{m} = 5\)
\((5)^2 = m^2 – 2 + \frac{1}{m^2}\)
Step 3: Simplify:
\(25 = m^2 + \frac{1}{m^2} – 2\)
\(m^2 + \frac{1}{m^2} = 25 + 2 = 27\)
Answer:27

ii. \(m^4+\frac{1}{m^4}\)

Step 1: Use identity: \((m^2+\frac{1}{m^2})^2 = m^4 + 2 + \frac{1}{m^4}\)
Step 2: Substitute \(m^2+\frac{1}{m^2} = 27\)
\((27)^2 = m^4 + 2 + \frac{1}{m^4}\)
Step 3: Simplify:
\(729 = m^4 + \frac{1}{m^4} + 2\)
\(m^4 + \frac{1}{m^4} = 729 – 2 = 727\)
Answer:727

iii. \(m^2-\frac{1}{m^2}\)

Step 1: Use identity: \((m-\frac{1}{m})^2 = m^2 – 2 + \frac{1}{m^2}\)
Step 2: \((m-\frac{1}{m})^2 + 4 = m^2 – \frac{1}{m^2})^2\) (identity: \(m^2 – \frac{1}{m^2} = (m-\frac{1}{m}) \cdot (m+\frac{1}{m})\))
Step 3: First, find \(m+\frac{1}{m} = \sqrt{(m-\frac{1}{m})^2 + 4} = \sqrt{25+4} = \sqrt{29}\)
Step 4: Then \(m^2 – \frac{1}{m^2} = (m-\frac{1}{m})(m+\frac{1}{m}) = 5 \cdot \sqrt{29} = 5\sqrt{29}\)
Answer:5√29

Q6: If \(a^2+b^2=41\) and \(ab=4\), find:

i. \(a-b\)

Step 1: Use identity: \((a-b)^2 = a^2 – 2ab + b^2)\)
Step 2: Substitute the given values:
\((a-b)^2 = 41 – 2(4) = 41 – 8 = 33\)
Step 3: Take the square root:
\(a-b = \sqrt{33}\)
Answer:\(\sqrt{33}\)

ii. \(a+b\)

Step 1: Use identity: \((a+b)^2 = a^2 + 2ab + b^2)\)
Step 2: Substitute the given values:
\((a+b)^2 = 41 + 2(4) = 41 + 8 = 49\)
Step 3: Take the square root:
\(a+b = \sqrt{49} = 7\)
Answer:7

Q7: If \(2a+\frac{1}{2a}=8\), find:

i. \(4a^2+\frac{1}{4a^2}\)

Step 1: Use the identity: \((x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}\)
Step 2: Here \(x = 2a\), so \(2a+\frac{1}{2a} = 8\)
\((2a+\frac{1}{2a})^2 = (2a)^2 + 2 + \frac{1}{(2a)^2}\)
Step 3: Substitute the value 8:
\(8^2 = 4a^2 + 2 + \frac{1}{4a^2}\)
\(64 = 4a^2 + \frac{1}{4a^2} + 2\)
\(4a^2 + \frac{1}{4a^2} = 64 – 2 = 62\)
Answer:62

ii. \(16a^4+\frac{1}{16a^4}\)

Step 1: Use the identity: \((x^2+\frac{1}{x^2})^2 = x^4 + 2 + \frac{1}{x^4}\)
Step 2: Substitute \(x^2+\frac{1}{x^2} = 4a^2+\frac{1}{4a^2} = 62\)
\((4a^2+\frac{1}{4a^2})^2 = (16a^4 + \frac{1}{16a^4}) + 2\)
\(62^2 = 16a^4 + \frac{1}{16a^4} + 2\)
Step 3: Simplify:
\(3844 = 16a^4 + \frac{1}{16a^4} + 2\)
\(16a^4 + \frac{1}{16a^4} = 3844 – 2 = 3842\)
Answer:3842

Q8: If \(3x-\frac{1}{3x}=5\), find:

i. \(9x^2+\frac{1}{9x^2}\)

Step 1: Use the identity: \((x-\frac{1}{x})^2 = x^2 – 2 + \frac{1}{x^2}\)
Step 2: Here \(x = 3x\), so \(3x-\frac{1}{3x} = 5\)
\((3x-\frac{1}{3x})^2 = (3x)^2 + \frac{1}{(3x)^2} – 2\)
Step 3: Substitute the value 5:
\(5^2 = 9x^2 + \frac{1}{9x^2} – 2\)
\(25 = 9x^2 + \frac{1}{9x^2} – 2\)
\(9x^2 + \frac{1}{9x^2} = 25 + 2 = 27\)
Answer:27

ii. \(81x^4+\frac{1}{81x^4}\)

Step 1: Use the identity: \((x^2+\frac{1}{x^2})^2 = x^4 + \frac{1}{x^4} + 2\)
Step 2: Substitute \(x^2+\frac{1}{x^2} = 9x^2+\frac{1}{9x^2} = 27\)
\((9x^2+\frac{1}{9x^2})^2 = 81x^4+\frac{1}{81x^4} + 2\)
\(27^2 = 81x^4+\frac{1}{81x^4} + 2\)
\(729 = 81x^4+\frac{1}{81x^4} + 2\)
\(81x^4+\frac{1}{81x^4} = 729 – 2 = 727\)
Answer:727

Q9: Expand:

i. \((3x-4y+5z)^2\)

Step 1: Use the identity: \((p+q+r)^2 = p^2 + q^2 + r^2 + 2pq + 2pr + 2qr\)
Step 2: Here \(p=3x, q=-4y, r=5z\)
\((3x-4y+5z)^2 = (3x)^2 + (-4y)^2 + (5z)^2 + 2(3x)(-4y) + 2(3x)(5z) + 2(-4y)(5z)\)
Step 3: Simplify each term:
\(= 9x^2 + 16y^2 + 25z^2 – 24xy + 30xz – 40yz\)
Answer:\(9x^2 + 16y^2 + 25z^2 – 24xy + 30xz – 40yz\)

ii. \((2a-5b-4c)^2\)

Step 1: \((p+q+r)^2 = p^2 + q^2 + r^2 + 2pq + 2pr + 2qr\)
Step 2: \(p=2a, q=-5b, r=-4c\)
\((2a-5b-4c)^2 = (2a)^2 + (-5b)^2 + (-4c)^2 + 2(2a)(-5b) + 2(2a)(-4c) + 2(-5b)(-4c)\)
Step 3: Simplify:
\(= 4a^2 + 25b^2 + 16c^2 – 20ab – 16ac + 40bc\)
Answer:\(4a^2 + 25b^2 + 16c^2 – 20ab – 16ac + 40bc\)

iii. \((5x+3y)^3\)

Step 1: Use the identity: \((p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3\)
Step 2: \(p=5x, q=3y\)
\((5x+3y)^3 = (5x)^3 + 3(5x)^2(3y) + 3(5x)(3y)^2 + (3y)^3\)
Step 3: Simplify:
\(= 125x^3 + 225x^2y + 135xy^2 + 27y^3\)
Answer:\(125x^3 + 225x^2y + 135xy^2 + 27y^3\)

iv. \((6a-7b)^3\)

Step 1: Use the identity: \((p-q)^3 = p^3 – 3p^2q + 3pq^2 – q^3\)
Step 2: \(p=6a, q=7b\)
\((6a-7b)^3 = (6a)^3 – 3(6a)^2(7b) + 3(6a)(7b)^2 – (7b)^3\)
Step 3: Simplify:
\(= 216a^3 – 756a^2b + 882ab^2 – 343b^3\)
Answer:\(216a^3 – 756a^2b + 882ab^2 – 343b^3\)

Q10: If \(a+b+c=9\) and \(ab+bc+ca=15\), find: \(a^2+b^2+c^2\).

Step 1: Use the identity: \((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
Step 2: Substitute the given values: \((9)^2 = a^2 + b^2 + c^2 + 2(15)\)
Step 3: Simplify: \(81 = a^2 + b^2 + c^2 + 30\)
Step 4: Solve for \(a^2 + b^2 + c^2\): \(a^2 + b^2 + c^2 = 81 – 30\)
Answer:51

Q11: If \(a+b+c=11\) and \(a^2+b^2+c^2=81\), find: \(ab+bc+ca\).

Step 1: Use the identity: \((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
Step 2: Substitute the given values:
\((11)^2 = 81 + 2(ab + bc + ca)\)
Step 3: Simplify: \(121 = 81 + 2(ab + bc + ca)\)
Step 4: Solve for \(ab + bc + ca\):
\(2(ab + bc + ca) = 121 – 81 = 40\)
\(ab + bc + ca = \frac{40}{2} = 20\)
Answer:20

Q12: If \(3x-4y=5\) and \(xy=3\), find: \(27x^3-64y^3\).

Step 1: Recognize the identity for difference of cubes: \(a^3-b^3=(a-b)\,(a^2+ab+b^2)\).
Step 2: Write \(27x^3-64y^3=(3x)^3-(4y)^3=(3x-4y)\big[(3x)^2+(3x)(4y)+(4y)^2\big]\).
Step 3: Compute the middle bracket using \(xy=3\): \((3x)(4y)=12xy=36\).
Step 4: Find \(9x^2+16y^2\) from \((3x-4y)^2=9x^2-24xy+16y^2\). Since \(3x-4y=5\) and \(xy=3\),
\(\;\;\;9x^2+16y^2=(3x-4y)^2+24xy=5^2+24\cdot3=25+72=97\).
Step 5: Therefore, the bracket equals \(9x^2+16y^2+12xy=97+36=133\).
Step 6: Hence \(27x^3-64y^3=(3x-4y)\times133=5\times133=665\).
Answer:665

Q13: If \(a+b=8\) and \(ab=15\), find: \(a^3+b^3\)

Step 1: Recall the sum of cubes identity: \(a^3+b^3 = (a+b)^3 – 3ab(a+b)\)
Step 2: Substitute the given values \(a+b=8\) and \(ab=15\): \(a^3+b^3 = (8)^3 – 3(15)(8)\)
Step 3: Simplify each term: \((8)^3 = 512\)
\(3*15*8 = 360\)
Step 4: Compute: \(a^3+b^3 = 512 – 360 = 152\)
Answer:152

Q14: If \(3x+2y=9\) and \(xy=3\), find: \(27x^3+8y^3\)

Step 1: Use sum of cubes: \(a^3+b^3=(a+b)\,(a^2-ab+b^2)\).
Step 2: Let \(a=3x,\; b=2y\). Then \(27x^3+8y^3=(3x)^3+(2y)^3=(3x+2y)\big[(3x)^2-(3x)(2y)+(2y)^2\big]\).
Step 3: We know \(3x+2y=9\) and \(xy=3\Rightarrow (3x)(2y)=6xy=18\).
Step 4: Find \(9x^2+4y^2\) from \((3x+2y)^2=9x^2+12xy+4y^2\). Thus \(9x^2+4y^2=9^2-12\cdot3=81-36=45\).
Step 5: Now the bracket is \(9x^2-6xy+4y^2=(9x^2+4y^2)-6xy=45-18=27\).
Step 6: Therefore \(27x^3+8y^3=(3x+2y)\times27=9\times27=243\).
Answer:243

Q15: If \(5x-4y=7\) and \(xy=8\), find: \(125x^3-64y^3\)

Step 1: Recognize the difference of cubes identity:
\((a^3 – b^3) = (a-b)(a^2 + ab + b^2)\)
Step 2: Express \(125x^3 – 64y^3\) as cubes:
\((5x)^3 – (4y)^3 = (5x – 4y) \left[(5x)^2 + (5x)(4y) + (4y)^2 \right ]\)
Step 3: Substitute the given values: \[ 5x – 4y = 7 \\ (5x)(4y) = 20xy = 20 \cdot 8 = 160 \\ (5x)^2 = 25x^2, \ (4y)^2 = 16y^2 \]Step 4: Compute \((5x)^2 + (5x)(4y) + (4y)^2\): \[ 25x^2 + 16y^2 + 160 \]Step 5: Express \(25x^2 + 16y^2\) in terms of \((5x-4y)^2\) and \(xy\): \[ (5x-4y)^2 = 25x^2 – 40xy + 16y^2 = 25x^2 – 320 + 16y^2 \\ \Rightarrow 25x^2 + 16y^2 = (5x-4y)^2 + 320 = 7^2 + 320 = 49 + 320 = 369 \]Step 6: Compute \(125x^3-64y^3\):
\(125x^3-64y^3 = (5x-4y)[(5x)^2 + (5x)(4y) + (4y)^2] = 7 \cdot (369 + 160) = 7 \cdot 529 = 3703\)
Answer:3703

Q16: The difference between two numbers is 5 and their product is 14. Find the difference between their cubes.

Step 1: Recall the identity for the difference of cubes:
\((a^3 – b^3) = (a-b)(a^2 + ab + b^2)\)
Step 2: Let the two numbers be \(a\) and \(b\). Given:
\(a-b = 5, \quad ab = 14\)
Step 3: Express \(a^2 + ab + b^2\) in terms of \(a-b\) and \(ab\): \[ (a-b)^2 = a^2 – 2ab + b^2 \Rightarrow a^2 + b^2 = (a-b)^2 + 2ab = 5^2 + 2 \cdot 14 = 25 + 28 = 53 \\ a^2 + ab + b^2 = a^2 + b^2 + ab = 53 + 14 = 67 \]Step 4: Compute the difference of cubes: \[ a^3 – b^3 = (a-b)(a^2 + ab + b^2) = 5 \cdot 67 = 335 \]Answer:335