If p = 101, then the value of $$\root 3 \of {p\left( {{p^2} - 3p + 3} \right) - 1} $$ is?
A. 100
B. 101
C. 102
D. 1000
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & p = 101 \cr & \root 3 \of {p\left( {{p^2} - 3p + 3} \right) - 1} \cr & = \root 3 \of {{p^3} - 3{p^2} + 3p - 1} \cr & \therefore \left[ {{{\left( {p - 1} \right)}^3} = {p^3} - {{\left( 1 \right)}^3} - 3p\left( {p - 1} \right)} \right] \cr & = \root 3 \of {{{\left( {p - 1} \right)}^3}} \cr & = p - 1 \cr & = 101 - 1 \cr & = 100{\text{ }} \cr} $$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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