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Question Number 56685 by pieroo last updated on 21/Mar/19

show that α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2}

Commented by pieroo last updated on 21/Mar/19

please, I need some help .

Answered by ajfour last updated on 22/Mar/19

α^{4} + β^{4} + 2 α^{2} β^{2} = {(α^{2} + β^{2})}^{2}

since

a^{2} + b^{2} + 2 ab = {(a + b)}^{2}

let a = α^{2} and b = β^{2}

\Rightarrow {(α^{2})}^{2} + {(β^{2})}^{2} + 2 (α^{2}) (β^{2}) = {(α^{2} + β^{2})}^{2} .

Commented by pieroo last updated on 22/Mar/19

thanks, but how did you get that ?

I know of the identity but tried severally

to prove it but to no avail .

Commented by MJS last updated on 23/Mar/19

{(α^{2} + β^{2})}^{2} = {(α^{2})}^{2} + 2 (α^{2}) (β^{2}) + {(β^{2})}^{2} = α^{4} + 2 α^{2} β^{2} + β^{4}

{where}^{'} s the problem ? ? ?

Commented by pieroo last updated on 28/Mar/19

alright alright . Thanks sir .

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