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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  −2α^2 β^2

$$\mathrm{show}\:\mathrm{that}\:\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} \:=\:\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)^{\mathrm{2}} \:−\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \\ $$

Commented by pieroo last updated on 21/Mar/19

please, I need some help.

$$\mathrm{please},\:\mathrm{I}\:\mathrm{need}\:\mathrm{some}\:\mathrm{help}. \\ $$

Answered by ajfour last updated on 22/Mar/19

α^4 +β^( 4) +2α^2 β^( 2) =(α^2 +β^( 2) )^2   since  a^2 +b^2 +2ab=(a+b)^2   let a=α^2  and b=β^( 2)   ⇒(α^2 )^2 +(β^( 2) )^2 +2(α^2 )(β^( 2) )=(α^2 +β^( 2) )^2  .

$$\alpha^{\mathrm{4}} +\beta^{\:\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{2}} \beta^{\:\mathrm{2}} =\left(\alpha^{\mathrm{2}} +\beta^{\:\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{since} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{2ab}=\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{a}=\alpha^{\mathrm{2}} \:\mathrm{and}\:\mathrm{b}=\beta^{\:\mathrm{2}} \\ $$$$\Rightarrow\left(\alpha^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\beta^{\:\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}\left(\alpha^{\mathrm{2}} \right)\left(\beta^{\:\mathrm{2}} \right)=\left(\alpha^{\mathrm{2}} +\beta^{\:\mathrm{2}} \right)^{\mathrm{2}} \:. \\ $$

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.

$$\mathrm{thanks},\:\mathrm{but}\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{that}? \\ $$$$\mathrm{I}\:\mathrm{know}\:\mathrm{of}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{but}\:\mathrm{tried}\:\mathrm{severally} \\ $$$$\mathrm{to}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{but}\:\mathrm{to}\:\mathrm{no}\:\mathrm{avail}. \\ $$

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 +2α^2 β^2 +β^4   where′s the problem???

$$\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)^{\mathrm{2}} =\left(\alpha^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}\left(\alpha^{\mathrm{2}} \right)\left(\beta^{\mathrm{2}} \right)+\left(\beta^{\mathrm{2}} \right)^{\mathrm{2}} =\alpha^{\mathrm{4}} +\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\beta^{\mathrm{4}} \\ $$$$\mathrm{where}'\mathrm{s}\:\mathrm{the}\:\mathrm{problem}??? \\ $$

Commented by pieroo last updated on 28/Mar/19

alright alright. Thanks sir.

$$\mathrm{alright}\:\mathrm{alright}.\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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