Question

(x+y+z) (x square + 9ysquare + 25z square -3xyz - 15yz - 5zx)
dont factorise
will mark branliest for good answer

Answer 1

Topic

• Polynomials- Identity

An identity is an equation that is true for all values of variables.

Proper Question

Expand $(x+3y+5z)(x^{2}+9y^{2}+25z^{2}-3xy-15yz-5zx)$.

Solution

Consider $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)=a^{3}+b^{3}+c^{3}-3abc$. The equation is true for all values of each variable. $^{\bold{[1]}}$

Thus, the equation above is an identity.

So, the required answer is the following.

$\implies x^{3}+(3y)^{3}+(5z)^{3}-3(x)(3y)(5z)=\boxed{x^{3}+9y^{3}+125z^{2}-45xyz}$.

More Information

What is an identity?

To solve an equation means finding the value of each variable that satisfies the equation. An identity is an equation that is true for all values of variables. So, solving an identity doesn't mean pretty much.

Information

$^{\bold{[1]}}$ To test if this is an identity, we can use basic concepts of polynomials, the distribution law.

Given Expression

$=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

But to calculate this, we need to use distribution law, which leads to calculating 18 terms.

So, let's take a different method. Let's divide $a^{3}+b^{3}+c^{3}-3abc$ by $a+b+c$.

By factor theorem, we obtain the following.

$\implies (-b-c)^{3}+b^{3}+c^{3}+3(b+c)bc$

However, this leads to the following.

$\implies (-b-c)^{3}+b^{3}+c^{3}+3(b+c)bc=0$

Hence the given expression is divisible by $a+b+c$.

By synthetic division method, we obtain the following.

$\implies a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

(Attachment included.)