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x-3-11-57-3-11-57-x-3-12x-




Question Number 10193 by konen last updated on 29/Jan/17
x=^3 (√(11+(√(57))))  +^3 (√(11−(√(57))))  ⇒ x^3 −12x=?
$$\mathrm{x}=^{\mathrm{3}} \sqrt{\mathrm{11}+\sqrt{\mathrm{57}}}\:\:+\:^{\mathrm{3}} \sqrt{\mathrm{11}−\sqrt{\mathrm{57}}} \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{3}} −\mathrm{12x}=? \\ $$
Answered by ridwan balatif last updated on 29/Jan/17
x^3 =(^3 (√(11+(√(57)))) +^3 (√(11−(√(57)))))^3   x^3 =(^3 (√(11+(√(57)))))^3 +(^3 (√(11−(√(57)))))^3 +3×(^3 (√(11+(√(57)))))(^3 (√(11−(√(57)))))(^3 (√(11+(√(57))))+^3 (√(11−(√(57)))))  remember: (a+b)^3 =a^3 +b^3 +3ab(a+b)  and also remember x=^3 (√(11+(√(57))))+^3 (√(11−(√(57)))),so  x^3 =11+(√(57)) +11−(√(57)) +3×(^3 (√(121−57)))(x)  x^3 =22+3x.(^3 (√(64)))  x^3 =22+12x  x^3 −12x=22
$$\mathrm{x}^{\mathrm{3}} =\left(^{\mathrm{3}} \sqrt{\mathrm{11}+\sqrt{\mathrm{57}}}\:+\:^{\mathrm{3}} \sqrt{\mathrm{11}−\sqrt{\mathrm{57}}}\right)^{\mathrm{3}} \\ $$$$\mathrm{x}^{\mathrm{3}} =\left(^{\mathrm{3}} \sqrt{\mathrm{11}+\sqrt{\mathrm{57}}}\right)^{\mathrm{3}} +\left(^{\mathrm{3}} \sqrt{\mathrm{11}−\sqrt{\mathrm{57}}}\right)^{\mathrm{3}} +\mathrm{3}×\left(^{\mathrm{3}} \sqrt{\mathrm{11}+\sqrt{\mathrm{57}}}\right)\left(^{\mathrm{3}} \sqrt{\mathrm{11}−\sqrt{\mathrm{57}}}\right)\left(^{\mathrm{3}} \sqrt{\mathrm{11}+\sqrt{\mathrm{57}}}+^{\mathrm{3}} \sqrt{\mathrm{11}−\sqrt{\mathrm{57}}}\right) \\ $$$$\mathrm{remember}:\:\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{3}} =\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{3ab}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{and}\:\mathrm{also}\:\mathrm{remember}\:\mathrm{x}=^{\mathrm{3}} \sqrt{\mathrm{11}+\sqrt{\mathrm{57}}}+^{\mathrm{3}} \sqrt{\mathrm{11}−\sqrt{\mathrm{57}}},\mathrm{so} \\ $$$$\mathrm{x}^{\mathrm{3}} =\mathrm{11}+\sqrt{\mathrm{57}}\:+\mathrm{11}−\sqrt{\mathrm{57}}\:+\mathrm{3}×\left(^{\mathrm{3}} \sqrt{\mathrm{121}−\mathrm{57}}\right)\left(\mathrm{x}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} =\mathrm{22}+\mathrm{3x}.\left(^{\mathrm{3}} \sqrt{\mathrm{64}}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} =\mathrm{22}+\mathrm{12x} \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{12x}=\mathrm{22} \\ $$

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