Question

plz answer with steps

Answer 1

Given : (x + y)^3 + (y + z)^3 + (z + x)^3 - 3(x + y)(y + z)(z + x)

We know that a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

= > (x + y + y + z + z + x)((x + y)^2 + (y + z)^2 + (z + x)^2 - (x + y)(y + z) - (y + z)(z + x) - (z + x)(x + y))

= > 2(x + y + z)((x^2 + y^2 + 2xy + y^2 + z^2 + 2yz + z^2 + x^2 + 2zx - xy - y^2 - xz - yz - yz - xy - z^2 - xz - xz - x^2 - yz - xy))

= > 2(x + y + z)(x^2 + y^2 + z^2 - xy - yz - xz)

= > 2(x^3 + y^3 + z^3 - x^2y - xyz - x^2z + x^2y + y^3 + z^2y - xy^2 - y^2z - xyz + x^2z + y^2z + z^3 - xyz - yz^2 - xz^2)

= > 2(x^3 + y^3 + z^3 - 3xyz).

Hope this helps!

Answer 2

In this question formula [a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c² - ab-bc-ca)] is used.

LHS part

(x+y)³ + (y+z)³ + (z+x)³ - 3(x+y)(y+z)(z+x)

=(x+y + y+z + z+x){(x+y)² + (y+z)² + (z+x)² - (x+y)(y+z) - (y+z)(z+x) - (z+x)(x+y)}

=2(x+y+z){x²+y²+2xy + y²+z²+2yz + z²+x²+2xz - xy-y²-xz-yz - yz-xy-z²-xz - xz-x²-yz-xy}

=2(x+y+z)(x² + y² + z² - xy - yz- zx)

2(x³+ y³+ z³ - 3xyz)

=RHS part

Proved

I think you forgot to include y in your question (3xyz)

Hope this helps :)