Question

8x³-1/27y³ Factorize it

Answer 1

Your Answer:

$\tt 8x^3 - \dfrac{1}{27y^3} \\\\ = \tt (2x)^3- (\dfrac{1}{3y})^3 \\\\ Now \ \ the \ \ following \ \ Question \ \ is \ \ in \ \ the \ \ form \\\\ \tt \star [a^3 - b^3 = (a-b)(a^2+b^2+ab)] \\\\ \tt So, \ \ taking \ \ the \ \ in \ \ Equation \ \ that \ \ form \\\\ \tt =(2x - \dfrac{1}{3y})[(2x)^2 + (\dfrac{1}{3y})^2 + (2x)(\dfrac{1}{3y})] \\\\ \tt = (2x-\dfrac{1}{3y})[4x^2 + \dfrac{1}{9y^2} + \dfrac{2x}{3y}]$

So, the factors are

$\tt (2x-\dfrac{1}{3y}) \ \ and \ \ [4x^2 + \dfrac{1}{9y^2} + \dfrac{2x}{3y}]$

Other Algebraic Identities:

$\tt \star (a - b)^2 = a^2 - 2ab + b^2 \\\\ \tt \star (a + b)^2 = a^2 + 2ab + b^2 \\\\ \tt \star a^2 - b^2= (a + b)(a - b) \\\\ \tt \star(x + a)(x + b) = x^2 + (a + b) x + ab \\\\ \tt \star (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \\\\ \tt \star (a + b)^3 = a^3 + b^3 + 3ab (a + b)$