Question Number 7912 by anshumansingh585@gmail.com last updated on 24/Sep/16
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =? \\ $$
Commented by prakash jain last updated on 24/Sep/16
$$\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right) \\ $$
Answered by ashis786 last updated on 24/Sep/16
$$\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$$ \\ $$
Answered by ashis786 last updated on 14/Dec/16
![a^3 +b^3 =(a+b)(a^2 −ab+b^2 ) =(a+b){a^2 +(ω+ω^2 )ab+ω^3 b^2 } [∵ ω+ω^2 =−1] [∵ ω^3 =1] =(a+b){a(a+bw)+ω^2 b(a+bw)} =(a+b)(a+bω)(a+bω^2 )](https://www.tinkutara.com/question/Q9545.png)
$$\:\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} \\ $$$$=\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right) \\ $$$$=\left({a}+{b}\right)\left\{{a}^{\mathrm{2}} +\left(\omega+\omega^{\mathrm{2}} \right){ab}+\omega^{\mathrm{3}} {b}^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\:\omega+\omega^{\mathrm{2}} =−\mathrm{1}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\:\omega^{\mathrm{3}} =\mathrm{1}\right] \\ $$$$=\left({a}+{b}\right)\left\{{a}\left({a}+{bw}\right)+\omega^{\mathrm{2}} {b}\left({a}+{bw}\right)\right\} \\ $$$$=\left({a}+{b}\right)\left({a}+{b}\omega\right)\left({a}+{b}\omega^{\mathrm{2}} \right) \\ $$