May 2026

Find a³ + b³ Question: If \[ a+b=8 \quad \text{and} \quad ab=6 \] find: \[ a^3+b^3 \] Solution: Using identity: \[ a^3+b^3=(a+b)^3-3ab(a+b) \] Substituting the given values: \[ a^3+b^3=(8)^3-3(6)(8) \] \[ =512-144 \] \[ =368 \] Next Question / Full Exercise

Find a² − ab + b² and a² + ab + b² Question: If \[ a+b=10 \quad \text{and} \quad ab=16 \] find the values of: \[ a^2-ab+b^2 \] and \[ a^2+ab+b^2 \] Solution: Using identity: \[ (a+b)^2=a^2+2ab+b^2 \] \[ 10^2=a^2+2(16)+b^2 \] \[ 100=a^2+32+b^2 \] \[ a^2+b^2=100-32 \] \[ a^2+b^2=68 \] Now, \[ a^2-ab+b^2 \] \[

Evaluate Using Identity Question: If \[ x=3 \] find the value of: \[ \left(\frac{5}{x} + 5x\right) \left(\frac{25}{x^2} – 25 + 25x^2\right) \] Solution: Rearranging the terms: \[ \left(\frac{5}{x} + 5x\right) \left(\frac{25}{x^2} – \frac{25x}{x} + 25x^2\right) \] Using identity: \[ (a+b)(a^2-ab+b^2)=a^3+b^3 \] Here, \[ a=\frac{5}{x},\qquad b=5x \] \[ = \left(\frac{5}{x}\right)^3 + (5x)^3 \] Substituting \[ x=3

Evaluate Using Identity Question: If \[ x=3 \quad \text{and} \quad y=-1 \] find the value of: \[ \left(\frac{x}{4} – \frac{y}{3}\right) \left(\frac{x^2}{16} + \frac{xy}{12} + \frac{y^2}{9}\right) \] Solution: Using identity: \[ (a-b)(a^2+ab+b^2)=a^3-b^3 \] Here, \[ a=\frac{x}{4},\qquad b=\frac{y}{3} \] \[ \left(\frac{x}{4} – \frac{y}{3}\right) \left(\frac{x^2}{16} + \frac{xy}{12} + \frac{y^2}{9}\right) \] \[ = \left(\frac{x}{4}\right)^3 – \left(\frac{y}{3}\right)^3 \] Substituting \[

Evaluate Using Identity Question: If \[ x=3 \quad \text{and} \quad y=-1 \] find the value of: \[ \left(\frac{x}{7} + \frac{y}{3}\right) \left(\frac{x^2}{49} + \frac{y^2}{9} – \frac{xy}{21}\right) \] Solution: Using identity: \[ (a+b)(a^2-ab+b^2)=a^3+b^3 \] Here, \[ a=\frac{x}{7},\qquad b=\frac{y}{3} \] \[ \left(\frac{x}{7} + \frac{y}{3}\right) \left(\frac{x^2}{49} – \frac{xy}{21} + \frac{y^2}{9}\right) \] \[ = \left(\frac{x}{7}\right)^3 + \left(\frac{y}{3}\right)^3 \] Substituting \[

Evaluate Using Identity Question: If \[ x=3 \] find the value of: \[ \left(\frac{3}{x} – \frac{x}{3}\right) \left(\frac{x^2}{9} + \frac{9}{x^2} + 1\right) \] Solution: Rearranging the terms: \[ \left(\frac{3}{x} – \frac{x}{3}\right) \left(\frac{9}{x^2} + \frac{3}{x}\cdot\frac{x}{3} + \frac{x^2}{9}\right) \] Using identity: \[ (a-b)(a^2+ab+b^2)=a^3-b^3 \] Here, \[ a=\frac{3}{x},\qquad b=\frac{x}{3} \] \[ = \left(\frac{3}{x}\right)^3 – \left(\frac{x}{3}\right)^3 \] Substituting \[ x=3

Evaluate Using Identity Question: If \[ x=3 \quad \text{and} \quad y=-1 \] find the value of: \[ (9y^2 – 4x^2)(81y^4 + 36x^2y^2 + 16x^4) \] Solution: Using identity: \[ (a-b)(a^2+ab+b^2)=a^3-b^3 \] Here, \[ a=9y^2,\qquad b=4x^2 \] \[ (9y^2 – 4x^2)(81y^4 + 36x^2y^2 + 16x^4) \] \[ = (9y^2)^3-(4x^2)^3 \] \[ = 729y^6-64x^6 \] Substituting \[

Find x³ − 1/x³ Question: \[ x – \frac{1}{x} = 3 + 2\sqrt{2} \] Find: \[ x^3 – \frac{1}{x^3} \] Solution: Using identity: \[ \left(a-b\right)^3 = a^3-b^3-3ab(a-b) \] \[ \left(x-\frac{1}{x}\right)^3 = x^3-\frac{1}{x^3} – 3\left(x\right)\left(\frac{1}{x}\right) \left(x-\frac{1}{x}\right) \] Since, \[ x\cdot \frac{1}{x}=1 \] \[ \left(x-\frac{1}{x}\right)^3 = x^3-\frac{1}{x^3} – 3\left(x-\frac{1}{x}\right) \] \[ x^3-\frac{1}{x^3} = \left(x-\frac{1}{x}\right)^3 + 3\left(x-\frac{1}{x}\right) \]

Find the Product (x³ + 1)(x⁶ − x³ + 1) Find the Product: \[ (x^3 + 1)(x^6 – x^3 + 1) \] Solution: Using identity: \[ (a+b)(a^2-ab+b^2)=a^3+b^3 \] Here, \[ a=x^3,\qquad b=1 \] \[ (x^3 + 1)(x^6 – x^3 + 1) \] \[ = (x^3 + 1)\left[(x^3)^2 – (x^3)(1) + 1^2\right] \] \[ = (x^3)^3

Find the Product (x² − 1)(x⁴ + x² + 1) Find the Product: \[ (x^2 – 1)(x^4 + x^2 + 1) \] Solution: Using identity: \[ (a-b)(a^2+ab+b^2)=a^3-b^3 \] Here, \[ a=x^2,\qquad b=1 \] \[ (x^2 – 1)(x^4 + x^2 + 1) \] \[ = (x^2 – 1)\left[(x^2)^2 + (x^2)(1) + 1^2\right] \] \[ = (x^2)^3