If $$\frac{1}{p} + \frac{1}{q}$$ = $$\frac{1}{{p + q}}{\text{,}}$$ then the value of p3 - q3 is?
A. p - q
B. pq
C. 1
D. 0
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & \frac{1}{p} + \frac{1}{q} = \frac{1}{{p + q}} \cr & \Rightarrow \frac{{p + q}}{{pq}} = \frac{1}{{p + q}} \cr & \Rightarrow {\left( {p + q} \right)^2} = pq \cr & \Rightarrow \left( {{p^2} + {q^2} + 2pq - pq} \right) = 0 \cr & \Rightarrow \left( {{p^2} + {q^2} + pq} \right) = 0 \cr & {\text{Multiply by }}\left( {p - q} \right){\text{ both side}} \cr & \Rightarrow \left( {p - q} \right)\left( {{p^2} + {q^2} + pq} \right) = \left( {p - q} \right) \times 0 \cr & \Rightarrow {p^3} - {q^3} = 0 \cr} $$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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