Question
Question: If \(x = 2 + {2^{\dfrac{2}{3}}} + {2^{\dfrac{1}{3}}}\), then the value of \({x^3} - 6{x^2} + 6x\,\)i...
If x=2+232+231, then the value of x3−6x2+6xis
A.3 B.2 C.1 D.−2
Solution
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+232+231 ……(i)
x3−6x2+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=a3−b3+3ab2−3a2b
On considering equation one and solving
x=2+232+231 x−2=232+231
Cubing both sides we get,
(x−2)3=(232+231)3 x3−8−6x2+12x=232(3)+231(3)+3.232(2).231+3.231(2).232
On subtracting 6x from both sides we get,
x3−6x2+12x−6x=8+6+3(235+234)−6x x3−6x2+6x=14+3(235+234)−6(2+232+231) x3−6x2+6x=2+3(235)+3(234)−6(232)−6(231) x3−6x2+6x=2+3(2)(232)+3(2)(231)−6(232)−6(231) x3−6x2+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=a3−b3+3ab2−3a2bin 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.