Question

Rationalize the denominator of 2/ ^3√a-1

Answer 1

do you know what a "conjugate" is?

Answer 2

Multiply top and bottom times, \(\large\color{blue}{ \rm \sqrt[3]{a-1} }\)

Answer 3

Sort of

Answer 4

\(\bf \Large \cfrac{2}{\sqrt[3]{a-1}}\quad ?\)

Answer 5

Is it basically the opposite?

Answer 6

Yes, that is the problem

Answer 7

so... haemm ok.... well what would you get if you multiply top and bottom by \(\bf \sqrt[3]{(a-1)^2}\) ?

Answer 8

1\[2\sqrt[3]{(a-1)^{2}}/a-1\]

Answer 9

I think

Answer 10

well one sec

Answer 11

\(\bf \large{ \cfrac{2}{\sqrt[3]{a-1}}\cdot \cfrac{\sqrt[3]{(a-1)^2}}{\sqrt[3]{(a-1)^2}}\implies \cfrac{2\sqrt[3]{(a-1)^2}}{\sqrt[3]{(a-1)^1}\cdot \sqrt[3]{(a-1)^2}}
\\ \quad \\
\cfrac{2\sqrt[3]{(a-1)^2}}{\sqrt[3]{(a-1)^{1+3}}}\implies \cfrac{2\sqrt[3]{(a-1)^2}}{\sqrt[3]{(a-1)^3}}
\\ \quad \\
\cfrac{2\sqrt[3]{(a-1)^2}}{a-1}
}\)

Answer 12

hmm \(\bf \large { \cfrac{2}{\sqrt[3]{a-1}}\cdot \cfrac{\sqrt[3]{(a-1)^2}}{\sqrt[3]{(a-1)^2}}\implies \cfrac{2\sqrt[3]{(a-1)^2}}{\sqrt[3]{(a-1)^1}\cdot \sqrt[3]{(a-1)^2}}
\\ \quad \\
\cfrac{2\sqrt[3]{(a-1)^2}}{\sqrt[3]{(a-1)^{1+2}}}\implies \cfrac{2\sqrt[3]{(a-1)^2}}{\sqrt[3]{(a-1)^3}}
\\ \quad \\
\cfrac{2\sqrt[3]{(a-1)^2}}{a-1} }\)

Answer 13

anyhow just a typo =)

Answer 14

the idea being, you'd multiply by a factor of the denominator, enough to get the radicand out

Answer 15

Thanks! I was a little confused on the radicand part on the bottom and if it just canceled out or what.