Question

if 3√x+(3√3375) ^3 =17 , then find the cube root of x-1026 is equal to​

Answer 1

SOLUTION :-

GIVEN :-

$\sf{ \sqrt[3]{x + { \bigg( \sqrt[3]{3375} \bigg)}^{3} } = 17 }$

TO DETERMINE :-

$\sf{ \sqrt[3]{x - 1026} }$

EVALUATION :-

Here it is given that

$\sf{ \sqrt[3]{x + { \bigg( \sqrt[3]{3375} \bigg)}^{3} } = 17 }$

Taking cube in both sides we get

$\sf{ {x + { \bigg( \sqrt[3]{3375} \bigg)}^{3} } = {(17)}^{3} }$

$\implies \sf{ {x + { \big( 15 \big)}^{3} } = 4913 }$

$\implies \sf{ x + 3375 = 4913 }$

$\implies \sf{ x = 4913 - 3375 }$

$\implies \sf{x = 1538}$

$\therefore \: \: \: \sf{ \sqrt[3]{x - 1026} }$

$= \sf{ \sqrt[3]{1538 - 1026} }$

$= \sf{ \sqrt[3]{512} }$

$= \sf{ \sqrt[3]{(8 \times 8 \times 8)} }$

$= \sf{8}$

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Answer 2

✮ Question

$\sf{If\:\:\sqrt[3]{x+\left(\sqrt[3]{3375}\right)^3}=17\:, Then\:find :\sqrt{x-1026}}$

✮ Answer  

$\sf{\sqrt{x-1026}=8}$

✮ Explanation

$\sf{\sqrt[3]{x+\left(\sqrt[3]{3375}\right)^3}=17}$

$\textsf { Take both sides of the equation to the power of 3}$

$\sf{\left(\sqrt[3]{x+\left(\sqrt[3]{3375}\right)^3}\right)^3=17^3}$

$\implies\sf{\left(\left(x+\left(\sqrt[3]{3375}\right)^3\right)^{\tfrac{1}{3}}\right)^3=4913}$

$\implies\sf{\left(x+\left(\sqrt[3]{3375}\right)^3\right)^{\tfrac{1}{3}\cdot \:3}=4913}$

$\implies\sf{x+\left(\sqrt[3]{3375}\right)^3=4913}$

$\sf{\implies x+3375=4913}$

$\implies\sf{x+3375-3375=4913-3375}$

$\implies\sf{x=1538}$

$\sf{Now \:\:Solve \:\:\sqrt[3]{x-1026} }$

$\implies\sf{\sqrt[3]{1538-1026}}$

$\implies\sf{\sqrt[3]{512}}$

$\implies\sf{\sqrt[3]{8^3}}$

$\Large\boxed{\implies\sf{8}}$