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# If $x = 2 + {2^{\dfrac{2}{3}}} + {2^{\dfrac{1}{3}}}$, then the value of ${x^3} - 6{x^2} + 6x\,$is$A.\,\,3 \\ B.\,\,2 \\ C.\,\,1 \\ D.\,\, - 2 \\$  Verified
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Hint : (Use the given equation to get the value of asked equation. Start solving from first equation and use that in second equation for final answer)
The given equations are
$x = 2 + {2^{\dfrac{2}{3}}} + {2^{\dfrac{1}{3}}}$ ……(i)
${x^3} - 6{x^2} + 6x\,$……(ii)
We have to find the value of second equation
We will do operation in first equation to get the value of second
As we know ${(a - b)^3} = {a^3} - {b^3} + 3a{b^2} - 3{a^2}b$
On considering equation one and solving
$x = 2 + {2^{\dfrac{2}{3}}} + {2^{\dfrac{1}{3}}} \\ x - 2 = {2^{\dfrac{2}{3}}} + {2^{\dfrac{1}{3}}} \\ \\$
Cubing both sides we get,
${(x - 2)^3} = {({2^{\dfrac{2}{3}}} + {2^{\dfrac{1}{3}}})^3} \\ {x^3} - 8 - 6{x^2} + 12x = {2^{\dfrac{{2(3)}}{3}}} + {2^{\dfrac{{1(3)}}{3}}} + {3.2^{\dfrac{{2(2)}}{3}}}{.2^{\dfrac{1}{3}}} + {3.2^{\dfrac{{1(2)}}{3}}}{.2^{\dfrac{2}{3}}} \\$
On subtracting $6x$ from both sides we get,
${x^3} - 6{x^2} + 12x - 6x = 8 + 6 + 3({2^{\dfrac{5}{3}}} + {2^{\dfrac{4}{3}}}) - 6x \\ {x^3} - 6{x^2} + 6x = 14 + 3({2^{\dfrac{5}{3}}} + {2^{\dfrac{4}{3}}}) - 6(2 + {2^{\dfrac{2}{3}}} + {2^{\dfrac{1}{3}}}) \\ {x^3} - 6{x^2} + 6x = 2 + 3({2^{\dfrac{5}{3}}}) + 3({2^{\dfrac{4}{3}}}) - 6({2^{\dfrac{2}{3}}}) - 6({2^{\dfrac{1}{3}}}) \\ {x^3} - 6{x^2} + 6x = 2 + 3(2)({2^{\dfrac{2}{3}}}) + 3(2)({2^{\dfrac{1}{3}}}) - 6({2^{\dfrac{2}{3}}}) - 6({2^{\dfrac{1}{3}}}) \\ {x^3} - 6{x^2} + 6x = 2 \\$

Hence the value of the second equation is 2 .
Therefore the correct option is B.

Note :- In this question we have used the formula of ${(a - b)^3} = {a^3} - {b^3} + 3a{b^2} - 3{a^2}b$in equation (i) above. We then got the value of equation (ii) by simplifying the equation (i). We have also used the concept of adding of the power when the base is same during obtaining the value of asked equation. There is nothing needed other than this to solve this question. Never try to put the values directly, it will make equation complex and the solution will not look good.