Question

2(x³+y³+z³-3xyz) how is this equation equal to 2(x³+y³+z³). i dont get it? where did 3xyz go​

Answer 1

Answer:

According to identity, we know that

(x+y+z)(x²+y²+z²-(xy+yz+zx))=x³+y³+z³-3xyz. Let this be equation 1.

But we don't know x²+y²+z².

We can find it by using

(x+y+z)²=x²+y²+z²+2(xy+yz+zx)

Substituting known values, we get

15²=x²+y²+z²+2(71)

x²+y²+z²=225-142

=83

Substituting known values and x²+y²+z² in equation 1, we get

15*(83-71)=x³+y³+z³-3(10)

x³+y³+z³=180-30

=150.

Step-by-step explanation:

thank you