Factorise the following expression: 27 x^{3}- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given expression is $27 x^{3}-8 y^{3}-54 x^{2} y+36 x y^{2}$.This can be rewritten as $(3x)^3 - (2y)^3 - 3(3x)^2(2y) + 3(3x)(2y)^2$.Recall the

Vedclass · Accepted Answer

$(3x - 2y)^3$

Vedclass · Answer

(A) The given expression is $27 x^{3}-8 y^{3}-54 x^{2} y+36 x y^{2}$.This can be rewritten as $(3x)^3 - (2y)^3 - 3(3x)^2(2y) + 3(3x)(2y)^2$.Recall the algebraic identity: $(a - b)^3 = a^3 - b^3 - 3a^2b + 3ab^2$.Here,$a = 3x$ and $b = 2y$.Substituting these values into the identity,we get $(3x - 2y)^3$.Thus,the factorised form is $(3x - 2y)(3x - 2y)(3x - 2y)$.

Factorise the following expression: $27 x^{3}-8 y^{3}-54 x^{2} y+36 x y^{2}$

Similar Questions

If $\frac{x}{y}+\frac{y}{x}=-1$ $(x, y \neq 0)$,the value of $x^{3}-y^{3}$ is

$\sqrt{2}$ is a polynomial of degree

Is $(x-1)$ a factor of $3x^2 + 7x - 10$?

Find the zero of the polynomial in the following case:
$p(x) = 3x - 4$

Check whether $(x-1)$ is a factor of the polynomial $p(x) = x^{3}-7x^{2}+14x-8$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module