Question

Find the value of x³+y³+z³-3xyz if x²+y²+z²=83 and x+y+z=15.

Please answer as fast as you can!!​

Answer 1

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.

Thus, 150 is the required answer.

Answer 2

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