If $$\frac{1}{{\root 3 \of 4 + \root 3 \of 2 + 1}}$$ = $$a\root 3 \of 4 $$ + $$b\root 3 \of 2 $$ + c and a, b, c are rational numbers then a + b + c is equal to?
A. 0
B. 1
C. 2
D. 3
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \frac{1}{{\root 3 \of 4 + \root 3 \of 2 + 1}} = a\root 3 \of 4 + b\root 3 \of 2 + c \cr & \Rightarrow \frac{1}{{\root 3 \of 4 + \root 3 \of 2 + 1}} = \frac{1}{{{{\left( {{2^{\frac{1}{3}}}} \right)}^2} + {2^{\frac{1}{3}}} + {{\left( 1 \right)}^2}}} \cr & \Rightarrow \therefore {{\text{A}}^3} - {{\text{B}}^3}{\text{ = }}\left( {{\text{A}} - {\text{B}}} \right)\left( {{{\text{A}}^2} + {\text{AB}} + {{\text{B}}^2}} \right) \cr & {\text{Put, A}} = {2^{\frac{1}{3}}}{\text{, B}} = 1 \cr & \Rightarrow \frac{{\left( {{2^{\frac{1}{3}}} - 1} \right)}}{{\left( {{2^{\frac{1}{3}}} - 1} \right)\left( {{{\left( {{2^{\frac{1}{3}}}} \right)}^2} + {2^{\frac{1}{3}}} + {{\left( 1 \right)}^2}} \right)}} \cr & \Rightarrow \frac{{\left( {{2^{\frac{1}{3}}} - 1} \right)}}{{{{\left( {{2^{\frac{1}{3}}}} \right)}^3} - {{\left( 1 \right)}^3}}} \cr & \Rightarrow \left( {{2^{\frac{1}{3}}} - 1} \right) \cr & \therefore {2^{\frac{1}{3}}} - 1 \cr & = a\left( {{2^{\frac{2}{3}}}} \right) + b{\left( 2 \right)^{\frac{1}{3}}} + c \cr & \left( {{\text{Comparing the terms}}} \right) \cr & a = 0 \cr & b = 1 \cr & c = - 1 \cr & \therefore a + b + c \cr & = 0 + 1 - 1 \cr & = 0 \cr} $$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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