Resolve into factors: 81 x^{2} y^{2}+108 x y | vedclass

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

Question

(C) The given expression is $81 x^{2} y^{2}+108 x y z+36 z^{2}$.We can rewrite this expression in the form of the algebraic identity $(a+b)^{2} = a^{2

Vedclass · Accepted Answer

$(9 x y+6 z)^{2}$

Vedclass · Answer

(C) The given expression is $81 x^{2} y^{2}+108 x y z+36 z^{2}$.We can rewrite this expression in the form of the algebraic identity $(a+b)^{2} = a^{2}+2ab+b^{2}$.Here,$a^{2} = (9 x y)^{2}$ and $b^{2} = (6 z)^{2}$.Thus,$a = 9 x y$ and $b = 6 z$.Now,check the middle term: $2ab = 2(9 x y)(6 z) = 108 x y z$.Since the expression matches the identity,we have:$81 x^{2} y^{2}+108 x y z+36 z^{2} = (9 x y+6 z)^{2}$.

Resolve into factors: $81 x^{2} y^{2}+108 x y z+36 z^{2}$.

Similar Questions

If $x+y+z=0,$ then what is the value of $\frac{x^{2}}{3yz} + \frac{y^{2}}{3xz} + \frac{z^{2}}{3xy}?$

Resolve into factors: $16 x^{2}-72 x y+81 y^{2}-12 x+27 y$

Resolve into factors: $16(x-y)^{2}-9(x+y)^{2}$

If $\frac{x}{y} + \frac{y}{x} = 6,$ then find the value of $\frac{x^{3}}{y^{3}} + \frac{y^{3}}{x^{3}}.$

If $x+y+z=0,$ what will be the value of $\frac{x^{2}+y^{2}+z^{2}}{x^{2}-y z} ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module