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steve jobs

18/03/2017 at 10:15
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if a+b = 4 and \(a^3+b^3=28,a^2+b^2=?\)

\([hint:\left(a+b\right)^2=a^2+2ab+b^2\)\(\left(a+b\right)^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)


Operation of Indices


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  • ...
    mathlove 18/03/2017 at 11:04

    By the assumption  \(\left\{{}\begin{matrix}a+b=4\\a^3+b^3=28\end{matrix}\right.\Rightarrow a^2-ab+b^2=\dfrac{28}{4}=7\Rightarrow2a^2-2ab+2b^2=14\).

    Other hand, \(a+b=4\Rightarrow a^2+2ab+b^2=16\). So:

               \(3\left(a^2+b^2\right)=\left(2a^2-2ab+2b^2\right)+\left(a^2+2ab+b^2\right)=14+16\Rightarrow a^2+b^2=10\)

    steve jobs selected this answer.
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    FA KAKALOTS 06/02/2018 at 12:30

    By the assumption  {a+b=4a3+b3=28⇒a2−ab+b2=284=7⇒2a2−2ab+2b2=14

    .

    Other hand, a+b=4⇒a2+2ab+b2=16

    . So:

               3(a2+b2)=(2a2−2ab+2b2)+(a2+2ab+b2)=14+16⇒a2+b2=10


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