If p = 124, then the value of $$\root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} = ?$$
A. 5
B. 7
C. 123
D. 125
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & p = 124 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {125} \right)}^3}} \cr & = 125 \cr} $$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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