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Given:
The objective is to solve the values of x for the equation,
[tex]x^5-36x^3=0[/tex]The value of x can be calculated by taking x to the power of 3 as common from both the terms.
[tex]\begin{gathered} x^3(x^2-36)=0 \\ (x^2-36)=0 \\ x^2=36 \\ x=\sqrt[]{36} \\ x=\pm6 \end{gathered}[/tex]Hence, the possibles values of x is +6, -6.
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Rewritten by : Barada
