Question

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Answer 1

Step-by-step explanation:

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Answer 2

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Consider the equation

x + y + z = 15

From algebraic identities,

we know that (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

So,

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

From the question,

x2 + y2 + z2 = 83 and x + y + z = 15

So,

152 = 83 + 2(xy + yz + xz)

=> 225 – 83 = 2(xy + yz + xz)

Or, xy + yz + xz = 142/2 = 71

Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),

x3 + y3 + z3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))

Now,

x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71

So, x3 + y3 + z3 – 3xyz = 15(83 – 71)

=> x3 + y3 + z3 – 3xyz = 15 × 12

Or, x3 + y3 + z3 – 3xyz = 180

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