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Question Number 187989 by mnjuly1970 last updated on 24/Feb/23

Answered by HeferH last updated on 24/Feb/23

$$b^2=ac$$$$b^3=abc$$$$\frac{b^3{c}^3+a^3{c}^3+a^3{b}^3}{a^3{b}^3{c}^3}\cdot \left(\frac{a^2{b}^2{c}^2}{a^3+b^3+c^3}\right)=$$$$\frac{b^3{c}^3+a^3{c}^3+a^3{b}^3}{abc}\cdot \left(\frac{1}{a^3+b^3+c^3}\right)=$$$$\frac{{\left(bc\right)}^3+b^6+{\left(ab\right)}^3}{b^3}\cdot \left(\frac{1}{a^3+b^3+c^3}\right)=$$$$c^3+b^3+a^3\cdot \left(\frac{1}{a^3+b^3+c^3}\right)=1$$$$$$

Answered by MathGuy last updated on 24/Feb/23

$$Answer:-$$$$\frac{a^2{b}^2{c}^2}{a^3+b^3+c^3}\left(\frac{b^3{c}^3+c^3{a}^3+a^3{b}^3}{a^3{b}^3{c}^3}\right)$$$$$$$$\Rightarrow \frac{1}{a^3+b^3+c^3}\left(\frac{b^3{c}^3+c^3{a}^3+a^3{b}^3}{abc}\right)$$$$$$$$\Rightarrow \frac{1}{a^3+b^3+c^3}(\frac{b^2{c}^2}{a}+\frac{c^2{a}^2}{b}+\frac{a^2{b}^2}{c})$$$$substitute{b}^2=ac\mathrm{\&}{c}^2{a}^2=b^2,intheexpression$$$$underbracketsandsimplify.$$$$youwillget$$$$\frac{1}{a^3+b^3+c^3}(a^3+b^3+c^3)$$$$=1$$$$$$$$$$

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