Find the value of x^{3}-8 y^{3}-36 x y-216,wh | vedclass

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

Question

(C) Given expression is $x^{3}-8 y^{3}-36 x y-216$.We can rewrite this as $x^{3} + (-2y)^{3} + (-6)^{3} - 3(x)(-2y)(-6)$.This is in the form of the al

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given expression is $x^{3}-8 y^{3}-36 x y-216$.We can rewrite this as $x^{3} + (-2y)^{3} + (-6)^{3} - 3(x)(-2y)(-6)$.This is in the form of the algebraic identity $a^{3} + b^{3} + c^{3} - 3abc = (a+b+c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$,where $a=x$,$b=-2y$,and $c=-6$.Substituting these values,we get: $(x - 2y - 6)(x^{2} + 4y^{2} + 36 + 2xy - 12y + 6x)$.Since we are given $x = 2y + 6$,it follows that $x - 2y - 6 = 0$.Therefore,the entire expression becomes $0 	imes (x^{2} + 4y^{2} + 36 + 2xy - 12y + 6x) = 0$.

Find the value of $x^{3}-8 y^{3}-36 x y-216$,when $x=2 y+6$.

Similar Questions

If $a+b+c=9$ and $ab+bc+ca=26$,find $a^{2}+b^{2}+c^{2}$.

Expand $(a - 2b + 7c)^2$.

Expand the following:
$(4a - b + 2c)^2$

Expand $\left(\frac{x}{2}-\frac{2}{5}\right)^{2}$.

Without finding the cubes,factorise $(x-y)^{3}+(y-z)^{3}+(z-x)^{3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module