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

$$\mathrm{show}\mathrm{that}{\alpha }^4+{\beta }^4={({\alpha }^2+{\beta }^2)}^2-2{\alpha }^2{\beta }^2$$

Commented by pieroo last updated on 21/Mar/19

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

Answered by ajfour last updated on 22/Mar/19

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

Commented by pieroo last updated on 22/Mar/19

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

Commented by MJS last updated on 23/Mar/19

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

Commented by pieroo last updated on 28/Mar/19

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

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