If x = $$2 - {2^{\frac{1}{3}}} + {2^{\frac{2}{3}}},$$ then find the value of x3 - 6x2 + 18x.
A. 40
B. 33
C. 45
D. 22
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & x = 2 - {2^{\frac{1}{3}}} + {2^{\frac{2}{3}}} \cr & x - 2 = {2^{\frac{2}{3}}} - {2^{\frac{1}{3}}} \cr & {\left( {x - 2} \right)^3} = {\left( {{2^{\frac{2}{3}}} - {2^{\frac{1}{3}}}} \right)^3} \cr & {x^3} - 8 - 3 \times 2x\left( {x - 2} \right) = 4 - 2 - 3 \times {2^{\frac{2}{3}}} \times {2^{\frac{1}{3}}}\left( {{2^{\frac{2}{3}}} - {2^{\frac{1}{3}}}} \right) \cr & {x^3} - 8 - 6{x^2} + 12x = 4 - 2 - 6\left( {x - 2} \right) \cr & {x^3} - 8 - 6{x^2} + 12x = 2 - 6x + 12 \cr & {x^3} - 6{x^2} + 12x + 6x = 2 + 12 + 8 \cr & {x^3} - 6{x^2} + 18x = 22 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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