The following problem is taken from available examples of Harvard University's entrance exam questions.
Given
\[ w^2-w^3=12 \]To find: \(w\).
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SOLUTION:
\[ w^2-w^3=12 \] \[ \Rightarrow w^3-w^2+12=0 \] \[ \Rightarrow w^3-w^2+8+4=0 \] \[ \Rightarrow (w^3+2^3)-(w^2-2^2)=0 \tag{1}\]Identity:
\begin{aligned} a^3+b^3&=(a+b)(a^2-ab+b^2) \\ a^2-b^2&=(a+b)(a-b) \\\\ \end{aligned} From (1): \[ (w+2)(w^2-2w+4)-(w+2)(w-2)=0 \] \[ \Rightarrow (w+2)(w^2-2w+4-w+2)=0 \] \[ \Rightarrow (w+2)(w^2-3w+6)=0 \]Therefore:
\[ w=\underline{-2} \] \[ w^2-3w+6=0 \tag{2} \]Quadratic formula:
\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2} \] From (2) : \[ w^2-3w+6=0\] \begin{aligned} \\\ \therefore w&=\frac{-(-3)\pm\sqrt{(-3)^2-4\cdot6}}{2} \\\\ &=\frac{+3\pm\sqrt{9-24}}{2} \\\\ &=\frac{+3\pm\sqrt{-15}}{2} \\\\ &=\underline{\frac{3\pm\sqrt{15}i}{2}} \\\\ \end{aligned}Solutions:
\[ \therefore \;\; w_1=\underline{\underline{-2}}, \;\; w_2=\underline{\underline{\frac{3+\sqrt{15}i}{2}}}, \;\; w_2=\underline{\underline{\frac{3-\sqrt{15}i}{2}}} \]