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Question Number 208469 by hardmath last updated on 16/Jun/24

$$\mathrm{Find}:\frac{{61}^3+{24}^3}{{61}^3+{37}^3}=?$$

Answered by Frix last updated on 16/Jun/24

$$a=61;b=24;a-b=37$$$$a^3+b^3=(a+b)(a^2-ab+b^2)$$$$a^3+{(a-b)}^3=$$$$=(a+(a-b))(a^2-a(a-b)+{(a-b)}^2)=$$$$=(2a-b)(a^2-ab+b^2)$$$$\Rightarrow \frac{a^3+b^3}{a^3+{(a-b)}^3}=\frac{a+b}{2a-b}$$$$\frac{61+24}{2\times 61-24}=\frac{85}{98}$$

Commented by hardmath last updated on 17/Jun/24

$$\mathrm{thankyou}\mathrm{dear}\mathrm{professor}$$

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