Question

square root of 10 - 2 root 12 is equal to root a + b root 3 then cube root of a minus b​

Answer 1

The value of $(a-b)^{^{\frac{1}{3}}$is $14^{^{\frac{1}{3}}$.

Step-by-step explanation:

We have,

$\sqrt{10-2\sqrt{12}} =\sqrt{a+b\sqrt{3}}$  ..... (1)

To find, $(a-b)^{^{\frac{1}{3}} =?$

Squaring (1) in both sides, we get

$10-2\sqrt{12}} ={a+b\sqrt{3}$

⇒ $10-2\sqrt{3\times 4}} ={a+b\sqrt{3}$

⇒$10-2\times 2\sqrt{3} ={a+b\sqrt{3}$

⇒ $10-4\sqrt{3} ={a+b\sqrt{3}$

Comparing both sides, we get

∴ a = 10 and $b\sqrt{3} = - 4\sqrt{3}$

b = - 4

∴ a - b = 10 - (- 4) = 14

$(a-b)^{^{\frac{1}{3}}=14^{^{\frac{1}{3}}$

Hence, the value of $(a-b)^{^{\frac{1}{3}}$is $14^{^{\frac{1}{3}}$.

Answer 2

Step-by-step explanation:

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