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Question Number 160744 by cortano last updated on 05/Dec/21

$${({\mathrm{x}}^2+\mathrm{x}-12)}^3+{({\mathrm{x}}^2+3\mathrm{x}-18)}^2=9{({\mathrm{x}}^2-9)}^2$$$$\mathrm{x}=?$$

Commented by blackmamba last updated on 05/Dec/21

$${(x+4)}^3{(x-3)}^3+{(x-3)}^2{(x+6)}^2=9{(x+3)}^2{(x-3)}^2$$$${(x-3)}^2\{(x-3){(x+4)}^3+{(x+6)}^2-9{(x+3)}^2\}=0$$$$\left(1\right)x=3$$$$\left(2\right)(x-3){(x+4)}^3+{(x+6)}^2-9{(x+3)}^2=0$$$$(x-3){(x+3+1)}^3+{(x+3+3)}^2-9{(x+3)}^2=0$$$$x+3=u$$$$(u-6){(u+1)}^3+{(u+3)}^2-9u^2=0$$$$(u-6)(u^3+3u^2+3u+1)+u^2+6u+9-9u^2=0$$$$...$$$$$$

Answered by MJS_new last updated on 06/Dec/21

$$\mathrm{obviously}x=3$$$${(x-3)}^3{(x+4)}^3+{(x-3)}^2{(x+6)}^2=9{(x-3)}^2{(x+3)}^2$$$$x_1=x_2=3$$$$(x-3){(x+4)}^3+{(x+6)}^2-9{(x+3)}^2=0$$$$x^4+9x^3+4x^2-122x-237=0$$$$\mathrm{this}\mathrm{has}\mathrm{got}\mathrm{no}‘‘{\mathrm{nice}}^{″}\mathrm{solution}$$$$x_3\approx -6.12906$$$$x_4\approx -3.74367$$$$x_5\approx -2.80700$$$$x_6\approx 3.67972$$

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