QUESTION IMAGE
Question
express in simplest radical form.
$sqrt3{24x^{3}y^{10}}$
Step1: Factor the radicand
We can factor \(24x^{3}y^{10}\) into parts that are perfect cubes (since we have a cube root).
- For the coefficient: \(24 = 8\times3\), and \(8 = 2^{3}\) (a perfect cube).
- For the \(x\)-term: \(x^{3}\) is already a perfect cube.
- For the \(y\)-term: \(y^{10}=y^{9}\times y=(y^{3})^{3}\times y\), where \(y^{9}\) is a perfect cube.
So, \(\sqrt[3]{24x^{3}y^{10}}=\sqrt[3]{8\times3\times x^{3}\times y^{9}\times y}\)
Step2: Apply the cube - root property \(\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}\)
Using the property of cube roots \(\sqrt[3]{abc}=\sqrt[3]{a}\times\sqrt[3]{b}\times\sqrt[3]{c}\), we can split the cube root:
\(\sqrt[3]{8\times3\times x^{3}\times y^{9}\times y}=\sqrt[3]{8}\times\sqrt[3]{x^{3}}\times\sqrt[3]{y^{9}}\times\sqrt[3]{3y}\)
Step3: Simplify each perfect - cube root
- We know that \(\sqrt[3]{8} = 2\) (because \(2^{3}=8\)).
- \(\sqrt[3]{x^{3}}=x\) (because \((x)^{3}=x^{3}\)).
- \(\sqrt[3]{y^{9}} = y^{3}\) (because \((y^{3})^{3}=y^{9}\)).
Substituting these values back, we get:
\(2\times x\times y^{3}\times\sqrt[3]{3y}\)
Step4: Combine the simplified terms
Multiplying the non - radical terms together, we have \(2xy^{3}\sqrt[3]{3y}\)
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\(2xy^{3}\sqrt[3]{3y}\)