steve jobs
18/03/2017 at 10:15-
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. -
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