\sqrt[3]{x} \left(x^{2} - 16\right) when x = -4/root(7)

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on Nov 26, 2024 at 6:24 am



You asked:

Calculate the value of x3(x2−16)\sqrt[3]{x} \left({x}^{2} - 16\right) when x=−472x = \frac{-4}{\sqrt[2]{7}}.

MathBot Answer:

x3(x2−16)=−96−13⋅223⋅75649\sqrt[3]{x} \left({x}^{2} - 16\right) = - \frac{96 \sqrt[3]{-1} \cdot 2^{\frac{2}{3}} \cdot 7^{\frac{5}{6}}}{49}

x3(x2−16)=−473((−47)2−16)=(−4)773(((−4)77)2−16)=(−4)773(167−16)=(167−16)49−13⋅223⋅756343=−13⋅223⋅7567(167−16)=(−96)−13⋅223⋅75649=−96−13⋅223⋅75649\begin{aligned}\sqrt[3]{x} \left(x^{2} - 16\right)&=\sqrt[3]{\frac{-4}{\sqrt{7}}} \left(\left(\frac{-4}{\sqrt{7}}\right)^{2} - 16\right)\\&=\sqrt[3]{\frac{\left(-4\right) \sqrt{7}}{7}} \left(\left(\frac{\left(-4\right) \sqrt{7}}{7}\right)^{2} - 16\right)\\&=\sqrt[3]{\frac{\left(-4\right) \sqrt{7}}{7}} \left(\frac{16}{7} - 16\right)\\&=\frac{\left(\frac{16}{7} - 16\right) 49 \sqrt[3]{-1} \cdot 2^{\frac{2}{3}} \cdot 7^{\frac{5}{6}}}{343}\\&=\frac{\sqrt[3]{-1} \cdot 2^{\frac{2}{3}} \cdot 7^{\frac{5}{6}}}{7} \left(\frac{16}{7} - 16\right)\\&=\frac{\left(-96\right) \sqrt[3]{-1} \cdot 2^{\frac{2}{3}} \cdot 7^{\frac{5}{6}}}{49}\\&=- \frac{96 \sqrt[3]{-1} \cdot 2^{\frac{2}{3}} \cdot 7^{\frac{5}{6}}}{49}\end{aligned}


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