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# help

0
80
2

If a^2 + 1/a^2 = 1, then find a^3 + 1/a^3.

May 18, 2020

#1
0

add 2, take a sqrt cube it and then expand

May 18, 2020
#2
+25555
+2

If
$$a^2 + \dfrac{1}{a^2}$$= 1,

then find
$$a^3 + \dfrac{1}{a^3}$$.

$$\begin{array}{|rcll|} \hline \left( a^2 + \dfrac{1}{a^2} \right) \left( a+\dfrac{1}{a} \right) &=& a^3+\dfrac{a^2}{a}+\dfrac{a}{a^2} + \dfrac{1}{a^3} \\\\ \left( a^2 + \dfrac{1}{a^2} \right) \left( a+\dfrac{1}{a} \right) &=& a^3+ \dfrac{1}{a^3}+ a +\dfrac{1}{a} \\\\ \left( a^2 + \dfrac{1}{a^2} \right) \left( a+\dfrac{1}{a} \right)-\left( a+\dfrac{1}{a} \right) &=& a^3+ \dfrac{1}{a^3} \\\\ \left( a+\dfrac{1}{a} \right) \left( a^2 + \dfrac{1}{a^2} -1 \right) &=& a^3+ \dfrac{1}{a^3} \quad | \quad a^2 + \dfrac{1}{a^2} = 1 \\\\ \left( a+\dfrac{1}{a} \right) \left(1 -1 \right) &=& a^3+ \dfrac{1}{a^3} \\\\ \left( a+\dfrac{1}{a} \right) \times 0 &=& a^3+ \dfrac{1}{a^3} \\\\ 0 &=& a^3+ \dfrac{1}{a^3} \\\\ \mathbf{ a^3+ \dfrac{1}{a^3} } &=& \mathbf{0} \\ \hline \end{array}$$

May 18, 2020