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Question Number 215730 by MathematicalUser2357 last updated on 16/Jan/25

$$\mathbf{Determine}\mathit{a},\mathit{b},\mathit{c}\left[\mathrm{Lazy}\mathrm{problem}\right]$$$$\mathrm{J}181-2.x^3-6x^2+15x-7={(x+a)}^3+bx+c$$$$\mathrm{J}182-\left(1\right)x^3+ax+2=(x+1)(x^2+bx+c)$$

Answered by A5T last updated on 16/Jan/25

$${(\mathrm{x}-2)}^3={\mathrm{x}}^3-6{\left(\mathrm{x}\right)}^2+12\mathrm{x}-8$$$$\Rightarrow {\mathrm{x}}^3-{6\mathrm{x}}^2+15\mathrm{x}-7={(\mathrm{x}-2)}^3+3\mathrm{x}+1$$$$\Rightarrow (\mathrm{a},\mathrm{b},\mathrm{c})=(-2,3,1)$$$$$$$${\mathrm{x}}^3+\mathrm{ax}+2={\mathrm{x}}^3+(\mathrm{b}+1){\mathrm{x}}^2+(\mathrm{b}+\mathrm{c})\mathrm{x}+\mathrm{c}$$$$\Rightarrow \mathrm{b}+1=0\Rightarrow \mathrm{b}=-1;\mathrm{c}=2;\mathrm{b}+\mathrm{c}=\mathrm{a}=2-1=1$$$$\Rightarrow (\mathrm{a},\mathrm{b},\mathrm{c})=(1,-1,2)$$

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