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Question Number 174800 by AgniMath last updated on 11/Aug/22
For which value of x this cubic equation will  be 0 ? a^3  − 16a − 3
$$\mathrm{For}\:\mathrm{which}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{this}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{0}\:?\:{a}^{\mathrm{3}} \:−\:\mathrm{16}{a}\:−\:\mathrm{3} \\ $$
Answered by MJS_new last updated on 11/Aug/22
a_1 =−((8(√3))/3)cos ((π/6)+(1/3)arcsin ((9(√3))/(128)))  a_2 =−((8(√3))/3)sin ((1/3)arcsin ((9(√3))/(128)))  a_3 =((8(√3))/3)sin ((π/3)+(1/3)arcsin ((9(√3))/(128)))
$${a}_{\mathrm{1}} =−\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{128}}\right) \\ $$$${a}_{\mathrm{2}} =−\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{128}}\right) \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arcsin}\:\frac{\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{128}}\right) \\ $$

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