Factorise x^{3}-8 y^{3}-27-18 x y. | vedclass

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

Question

(B) The given expression is $x^{3}-8 y^{3}-27-18 x y$.This can be rewritten as $x^{3} + (-2y)^{3} + (-3)^{3} - 3(x)(-2y)(-3)$.We use the algebraic ide

Vedclass · Accepted Answer

$(x-2y-3)(x^2+4y^2+9+2xy-6y+3x)$

Vedclass · Answer

(B) The given expression is $x^{3}-8 y^{3}-27-18 x y$.This can be rewritten as $x^{3} + (-2y)^{3} + (-3)^{3} - 3(x)(-2y)(-3)$.We use the algebraic identity $a^{3} + b^{3} + c^{3} - 3abc = (a+b+c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.Here,$a = x$,$b = -2y$,and $c = -3$.Substituting these values into the identity:$= (x - 2y - 3)(x^{2} + (-2y)^{2} + (-3)^{2} - (x)(-2y) - (-2y)(-3) - (-3)(x))$$= (x - 2y - 3)(x^{2} + 4y^{2} + 9 + 2xy - 6y + 3x)$.

Factorise $x^{3}-8 y^{3}-27-18 x y$.

Similar Questions

$85 \times 75 = \ldots \ldots \ldots$

Simplify $(2 x-5 y)^{3}-(2 x+5 y)^{3}$

Verify whether $2$ and $5$ are zeros of the polynomial $x^{2}-2x-15$ or not.

Factorise $: x^{3}-125$

If $p(x) = x + 3,$ then $p(x) + p(-x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module