Question

Evaluate (77-20√13)^(1÷3) + (77+20√13)^(1÷3)

Answer 1

Answer:

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Step-by-step explanation:

(77-20√13)^(1÷3) + (77+20√13)^(1÷3) =7

Answer 2

Answer:

The value of $\big(77-20\sqrt{13} \big)^{1/3} +\big (77+20\sqrt{13}\big)^{1/3}=7$

Step-by-step explanation:

Given to evaluate

$\big(77-20\sqrt{13} \big)^{1/3} +\big (77+20\sqrt{13}\big)^{1/3}$

Let $a=\big(77-20\sqrt{13} \big)^{1/3} ~~\mbox{and}~~b=\big (77+20\sqrt{13}\big)^{1/3}$

Taking cube on both sides for both a and b, we get

$a^3=\big(77-20\sqrt{13} \big)^{3/3} ~~\mbox{and}~~b^3=\big (77+20\sqrt{13}\big)^{3/3}$

$\implies a^3=77-20\sqrt{13} ~~\mbox{and}~~b^3=77+20\sqrt{13}\big$

Adding both the expressions,

$a^3+b^3=77-20\sqrt{13} +77+20\sqrt{13}\big$

$\implies a^3+b^3=154$                                                         ...(1)

Now taking product of a and b,

$ab=\big(77-20\sqrt{13} \big)^{1/3} \big (77+20\sqrt{13}\big)^{1/3}$

$ab=\Big[\big(77-20\sqrt{13} \big) \big (77+20\sqrt{13}\big)\Big]^{1/3}$

Using the algebraic identity,

$a^{2} -b^{2} =(a+b)(a-b)$

$ab=\Big[\big(77\big)^2-\big(20\sqrt{13} \big)^2\Big]^{1/3}$

   $=\big(5929-5200\big)^{1/3}=\big(729\big)^{1/3}$

   $=\big(9\big)^{3/3}=9$

Therefore, $ab=9$                                                           ...(2)

Now using the other algebraic identity,

$(a+b)^3=a^{3} +b^{3} +3ab(a+b)$

Substituting equation (1) and (2), and taking $a+b=x$

$(x)^3=154 +3\times 9(x)$

$\implies x^3 -27x-154=0$                                                ...(3)

The factors of 154 are 1, 7 and 11,

then the possible roots may be $\pm1, \pm7, \pm11$.

By substituting each value if we check the equation (3), only 7 satisfies the equation.

$\implies 7^3 -27\times 7-154=0$

Therefore, 7 is the root of the equation.

Since $a+b=x\implies a+b=7$

Therefore, the value of $\big(77-20\sqrt{13} \big)^{1/3} +\big (77+20\sqrt{13}\big)^{1/3}=7$.

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