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Oksanka [162]
2 years ago
8

5B3%5D%7B27y%5E2%7D" id="TexFormula1" title="\sqrt[3]{y} (7\sqrt[3]{8y^2}-\sqrt[3]{y^5} -4y\sqrt[3]{27y^2}" alt="\sqrt[3]{y} (7\sqrt[3]{8y^2}-\sqrt[3]{y^5} -4y\sqrt[3]{27y^2}" align="absmiddle" class="latex-formula"> simplify
Mathematics
1 answer:
noname [10]2 years ago
7 0

Answer:

\huge\boxed{-y^2+2y}

Step-by-step explanation:

\sqrt[3]y\cdot\left(7\sqrt[3]{8y^2}-\sqrt[3]{y^5}-4y\sqrt[3]{27y^2}\right)\\\\=(\sqrt[3]y)(7\sqrt[3]{8y^2})-(\sqrt[3]y)(\sqrt[3]{y^5})-(\sqrt[3]y)(4y\sqrt[3]{27y^2})\\\\=7\sqrt[3]{(y)(8y^2)}}-\sqrt[3]{(y)(y^5)}-4y\sqrt[3]{(y)(27y^2)}\\\\=7\sqrt{8y^3}-\sqrt{y^6}-4\sqrt{27y^3}\\\\=7\sqrt[3]{2^3y^3}-\sqrt{y^{2\cdot3}}-4\sqrt{3^3y^3}\\\\=7\sqrt[3]{(2y)^3}-\sqrt{(y^2)^3}-4\sqrt{(3y)^3}\\\\=7\cdot2y-y^2-4\cdot3y\\\\=14y-y^2-12y\\\\=-y^2+2y

Used:

a(a+b)=ab+ac\\\\\sqrt[3]{a\cdot b}=\sqrt[3]a\cdot\sqrt[3]b\\\\\sqrt[3]{a^3}=a\\\\(a^n)^m=a^{n\cdot m}

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