Given a - b = 2, a3 - b3 = 26, then (a + b)2 is?
A. 9
B. 4
C. 16
D. 12
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & a - b = 2{\text{ }} \cr & {a^3} - {b^3} = 26 \cr & \Rightarrow {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 26 = \left( 2 \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 13 = \left( {{a^2} + ab + {b^2}} \right)\,....(i) \cr & \Rightarrow 4 = 13 + ab \cr & \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \cr & \Rightarrow {\left( 2 \right)^2} = {a^2} + {b^2} + ab - 3ab \cr & \Rightarrow 3ab = 9 \cr & \Rightarrow ab = 3 \cr & \therefore {\left( {a + b} \right)^2} \cr & = {\left( {a - b} \right)^2} + 4ab \cr & = 4 + 4 \times 3 \cr & = 16 \cr} $$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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