Х Given a + b +c+d=0, prove that a3 + b3 + c3 + d3 = 3 (abc + bcd + cda + dab).
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this is it . Hope you are happy.
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Step-by-step explanation:
Given,
a+b+c+d=0
⇒(a+b)= -(c+d)
⇒(a+b)³=[-(c+d)]³
⇒a³+b³+3ab(a+b)= -[c³+d³+3cd(c+d)]
⇒a³+b³+c³+d³= -(c+d)3cd-(a+b)3ab
⇒a³+b³+c³+d³=(a+b)3cd+(c+d)3ab
⇒a³+b³+c³+d³=3(abc + bcd + cda + dab) [proved]
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