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Given a - b = 2, a³ - b³ = 26, then (a + b)² is?

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a - b = 2 \\ a^{3} - b^{3} = 26 \\ \Rightarrow a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \\ \Rightarrow 26 = (2) (a^{2} + a b + b^{2}) \\ \Rightarrow 13 = (a^{2} + a b + b^{2}) . . . . (i) \\ \Rightarrow 4 = 13 + a b \\ \Rightarrow {(a - b)}^{2} = a^{2} + b^{2} - 2 a b \\ \Rightarrow {(2)}^{2} = a^{2} + b^{2} + a b - 3 a b \\ \Rightarrow 3 a b = 9 \\ \Rightarrow a b = 3 \\ ∴ {(a + b)}^{2} \\ = {(a - b)}^{2} + 4 a b \\ = 4 + 4 \times 3 \\ = 16 \end{aligned}

$\eqalign{ & a - b = 2{\text{ }} \cr & {a^3} - {b^3} = 26 \cr & \Rightarrow {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 26 = \left( 2 \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 13 = \left( {{a^2} + ab + {b^2}} \right)\,....(i) \cr & \Rightarrow 4 = 13 + ab \cr & \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \cr & \Rightarrow {\left( 2 \right)^2} = {a^2} + {b^2} + ab - 3ab \cr & \Rightarrow 3ab = 9 \cr & \Rightarrow ab = 3 \cr & \therefore {\left( {a + b} \right)^2} \cr & = {\left( {a - b} \right)^2} + 4ab \cr & = 4 + 4 \times 3 \cr & = 16 \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a - b = 2 \\ a^{3} - b^{3} = 26 \\ \Rightarrow a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \\ \Rightarrow 26 = (2) (a^{2} + a b + b^{2}) \\ \Rightarrow 13 = (a^{2} + a b + b^{2}) . . . . (i) \\ \Rightarrow 4 = 13 + a b \\ \Rightarrow {(a - b)}^{2} = a^{2} + b^{2} - 2 a b \\ \Rightarrow {(2)}^{2} = a^{2} + b^{2} + a b - 3 a b \\ \Rightarrow 3 a b = 9 \\ \Rightarrow a b = 3 \\ ∴ {(a + b)}^{2} \\ = {(a - b)}^{2} + 4 a b \\ = 4 + 4 \times 3 \\ = 16 \end{aligned}

$\eqalign{ & a - b = 2{\text{ }} \cr & {a^3} - {b^3} = 26 \cr & \Rightarrow {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 26 = \left( 2 \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 13 = \left( {{a^2} + ab + {b^2}} \right)\,....(i) \cr & \Rightarrow 4 = 13 + ab \cr & \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \cr & \Rightarrow {\left( 2 \right)^2} = {a^2} + {b^2} + ab - 3ab \cr & \Rightarrow 3ab = 9 \cr & \Rightarrow ab = 3 \cr & \therefore {\left( {a + b} \right)^2} \cr & = {\left( {a - b} \right)^2} + 4ab \cr & = 4 + 4 \times 3 \cr & = 16 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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