A+A'=I find the value of x
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(d) Now, a3+b3 + c3 = (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc
[using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)] = 0 + 3abc [∴ a + b + c = 0]
a3+b3 + c3 = 3abc
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