Question

question
rewrite \\(\sqrt3{j^{25}}\\) in simplest form.
answer attempt 1 out of 2

Explanation:

Step1: Recall the property of radicals and exponents: $\sqrt[3]{a^b}=a^{\frac{b}{3}}$, and also use the division of exponents with the same base (when expressing in terms of quotient and remainder). We need to divide the exponent 25 by 3.

Divide 25 by 3: $25 = 3\times8 + 1$, so $j^{25}=j^{3\times8 + 1}=(j^{3})^8\times j^{1}$

Step2: Apply the cube - root property $\sqrt[3]{a\times b}=\sqrt[3]{a}\times\sqrt[3]{b}$ and $\sqrt[3]{a^n}=a^{\frac{n}{3}}$ (when n is a multiple of 3).

$\sqrt[3]{j^{25}}=\sqrt[3]{(j^{3})^8\times j}=\sqrt[3]{(j^{3})^8}\times\sqrt[3]{j}$
Since $\sqrt[3]{(j^{3})^8}=j^{8}$ (because $\sqrt[3]{a^{3k}} = a^{k}$ when k is an integer, here $a = j$ and $3k=24$ so $k = 8$) and $\sqrt[3]{j}=j^{\frac{1}{3}}$
So $\sqrt[3]{j^{25}}=j^{8}\sqrt[3]{j}$

Answer:

$j^{8}\sqrt[3]{j}$