Question

(3a – 1/b)^3 – 6(3a – 1/b) + 9 + (c + 1/b – 2a) (3a – 1/b – 3) Solve this​

Answer 1

Answer:

Correct Question:

$( 3a - \frac{1}{b} ) {}^{3} - 6(3a - \frac{1}{b} ) + 9 + \\ (c + \frac{1}{b} - 2a)(3a - \frac{1}{b} - 3)$

$= (3a - \frac{1}{b} ) {}^{2} - 2 \times 3(3a - \frac{1}{b} ) \\ + 3 {}^{2} + (c + \frac{1}{b} - 2a)(3a - \frac{1}{b} - 3)$

$= (3a - \frac{1}{b} - 3) (3a - \frac{b}{3} - 3) + \\ (c + \frac{1}{b} - 2a)(3a - \frac{1}{b} - 3)$

$= (3a - \frac{1}{b} - 3)(3a - \frac{1}{b} - 3 + c + \frac{1}{b} - 2a)$

$= (3a - \frac{1}{b} - 3)(a + c - 3)$

Hence, (3a – 1/b – 3)(a + c – 3) is the solution.

Some algebraic identities

→ (a + b)^2 = a^2 + 2ab + b^2

→ (a – b)^2 = a^2 – 2ab + b^2

→ a^2 – b^2 = (a + b) (a – b)

→ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

→ (a + b – c)^2 = a^2 + b^2 + c^2 + 2ab – 2bc – 2ca

→ (a – b – c)^2 = a^2 + b^2 + c^2 – 2ab + 2bc – 2ca

→ (a + b)^3 = a^3 + b^3 + 3ab(a + b)

→ (a – b)^3 = a^3 – b^3 – 3ab(a – b)

→ (a^3 + b^3) = (a + b) (a^2 – ab + b^2)

→ (a^3 – b^3) = (a – b) (a^2 + ab + b^2)