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Question Number 205884 by Abdullahrussell last updated on 01/Apr/24

Commented by Frix last updated on 01/Apr/24

$$a^3+b^3$$

Answered by Rasheed.Sindhi last updated on 04/Apr/24

$$\frac{{(a^4+2{ab}^3)}^3-{({ab}^4+2a^{3}b)}^3}{{(a^3-b^3)}^3}$$$$=\frac{{\mathrm{A}}^3-{\mathrm{B}}^3}{{(a^3-b^3)}^3}$$$$=\frac{(\mathrm{A}-\mathrm{B})({(\mathrm{A}-\mathrm{B})}^2+3\mathrm{AB})}{{(a-b)}^3{(a^2+ab+b^2)}^3}$$$$=\frac{\mathrm{A}-\mathrm{B}}{{(a-b)}^3{(a^2+ab+b^2)}^3}\cdot ({(\mathrm{A}-\mathrm{B})}^2+3\mathrm{AB})$$$$\mathrm{A}-\mathrm{B}=a^4-b^4+2{ab}^3-2a^{3}b$$$$=(a^2-b^2)(a^2+b^2)-2ab(a^2-b^2)$$$$=(a^2-b^2)(a^2+b^2-2ab)$$$$={(a-b)}^3(a+b)$$$$\frac{\mathrm{A}-\mathrm{B}}{{(a-b)}^3{(a^2+ab+b^2)}^3}=\frac{\cancel{{(a-b)}^3}(a+b)}{\cancel{{(a-b)}^3}{(a^2+ab+b^2)}^3}$$$$=\frac{a+b}{{(a^2+ab+b^2)}^3}$$$$({(\mathrm{A}-\mathrm{B})}^2+3\mathrm{AB})$$$$=({\left({(a-b)}^3(a+b)\right)}^2+3(a^4+2{ab}^3)(b^4+2a^{3}b))$$$$=({(a-b)}^6{(a+b)}^2+3(a^4{b}^4+4a^4{b}^4+2{ab}^7+2a^{7}b))$$$$=({(a-b)}^6{(a+b)}^2+15a^4{b}^4+6ab(a^6+b^6))$$$$=(a^2-ab+b^2){(a^2+ab+b^2)}^3$$$$\frac{\mathrm{A}-\mathrm{B}}{{(a-b)}^3{(a^2+ab+b^2)}^3}\cdot ({(\mathrm{A}-\mathrm{B})}^2+3\mathrm{AB})$$$$=\frac{a+b}{\cancel{{(a^2+ab+b^2)}^3}}\cdot (a^2-ab+b^2)\cancel{{(a^2+ab+b^2)}^3}$$$$=(a+b)(a^2-ab+b^2)=a^3+b^3$$

Answered by A5T last updated on 03/Apr/24

$$=\frac{a^{12}+8a^3{b}^9+6a^9{b}^3-b^{12}-8a^9{b}^3-6a^3{b}^9}{{(a^3-b^3)}^3}$$$$=\frac{a^{12}-b^{12}+2a^3{b}^9-2a^9{b}^3}{{(a^3-b^3)}^3}=\frac{(a^6-b^6)(a^6+b^6)-2a^3{b}^3(-b^6+a^6)}{{(a^3-b^3)}^3}$$$$=\frac{(a^3+b^3)(a^3-b^3)(a^6+b^6-2a^3{b}^3)}{{(a^3-b^3)}^3}=a^3+b^3$$

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