Question

81x^{15}y^{10}z^{9} ; 18x^{8}y^{7}z^{10}

Explanation:

Step1: Analyze the expression \(81x^{15}y^{10}z^{9}\)

We can rewrite \(81\) as \(3^4\), \(x^{15}\) as \((x^3)^5\) (wait, actually, to find the ninth root? Wait, maybe it's the fourth root? Wait, no, let's check the exponents. Wait, maybe we are taking the fourth root? Wait, \(81 = 3^4\), \(x^{15}\), \(y^{10}\), \(z^9\). Wait, maybe the problem is to simplify \(\sqrt[4]{81x^{15}y^{10}z^{9}}\)? Let's assume that. So first, simplify each part:

For the coefficient: \(\sqrt[4]{81}=\sqrt[4]{3^4} = 3\)

For \(x^{15}\): \(\sqrt[4]{x^{15}}=x^{\frac{15}{4}}=x^{3+\frac{3}{4}} = x^3x^{\frac{3}{4}}\) or maybe we can write it as \(x^3\sqrt[4]{x^3}\)

For \(y^{10}\): \(\sqrt[4]{y^{10}}=y^{\frac{10}{4}}=y^{2+\frac{2}{4}}=y^2\sqrt[4]{y^2}=y^2\sqrt{y}\)

For \(z^9\): \(\sqrt[4]{z^9}=z^{2+\frac{1}{4}}=z^2\sqrt[4]{z}\)

Wait, maybe the original problem is to take the fourth root? Wait, the user's image shows \(81x^{15}y^{10}z^9\) and maybe another expression. Wait, maybe the problem is to simplify \(\sqrt[4]{81x^{15}y^{10}z^9}\). Let's do that step by step.

Step1: Simplify the coefficient

\(\sqrt[4]{81}=\sqrt[4]{3^4} = 3\)

Step2: Simplify \(x^{15}\) under the fourth root

\(\sqrt[4]{x^{15}}=\sqrt[4]{x^{12}\cdot x^3}=x^3\sqrt[4]{x^3}\) (since \(x^{12}=(x^3)^4\))

Step3: Simplify \(y^{10}\) under the fourth root

\(\sqrt[4]{y^{10}}=\sqrt[4]{y^8\cdot y^2}=y^2\sqrt[4]{y^2}=y^2\sqrt{y}\) (since \(y^8=(y^2)^4\))

Step4: Simplify \(z^9\) under the fourth root

\(\sqrt[4]{z^9}=\sqrt[4]{z^8\cdot z}=z^2\sqrt[4]{z}\) (since \(z^8=(z^2)^4\))

Step5: Combine all parts

Putting it together, \(\sqrt[4]{81x^{15}y^{10}z^9}=3x^3y^2z^2\sqrt[4]{x^3y^2z}\)

Wait, maybe the problem is different. Alternatively, maybe it's a typo and the exponent for z is 4? No, the image shows \(z^9\). Alternatively, maybe the problem is to factor or simplify the expression. Wait, another possibility: \(81x^{15}y^{10}z^9 = 3^4x^{15}y^{10}z^9\). Maybe we can factor into perfect fourth powers?

\(x^{15}=x^{12}\cdot x^3=(x^3)^4\cdot x^3\)

\(y^{10}=y^8\cdot y^2=(y^2)^4\cdot y^2\)

\(z^9=z^8\cdot z=(z^2)^4\cdot z\)

So \(81x^{15}y^{10}z^9 = 3^4(x^3)^4(y^2)^4(z^2)^4\cdot x^3y^2z=(3x^3y^2z^2)^4\cdot x^3y^2z\)

Therefore, \(\sqrt[4]{81x^{15}y^{10}z^9}=3x^3y^2z^2\sqrt[4]{x^3y^2z}\)

Alternatively, if the problem is to write as a perfect fourth power times a remainder, that's what we did.

Answer:

If simplifying \(\sqrt[4]{81x^{15}y^{10}z^9}\), the result is \(3x^3y^2z^2\sqrt[4]{x^3y^2z}\)