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Question

If a + b + c = 8 and ab + bc + ca = 20, find the value of a3+b3+c33abc.

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Solution

Since (a+b+c)2=a2+b2+c2+2(ab+bc+ca)

a+b+c=8 and ab+bc+ca=20

(8)2=a2+b2+c2+2×20

64=a2+b2+c2+40

a2+b2+c2=24

We now use the following identity:
a3+b3+c33abc=(a+b+c)(a2+b2+c2(ab+bc+ca))

a3+b3+c33abc=8×(2420)=4×8=32

Thus, a3+b3+c33abc=32

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