Question

$a \sqrt{a} + b \sqrt{b} = 183 \\ a \sqrt{b} + b \sqrt{a} = 182 \\ then \: find \: the \: value \: of \: (a + b)$

Answer 1

$a \sqrt{a} + b \sqrt{b} = 183 \\ {( \sqrt{a}) }^{3} + {( \sqrt{b}) }^{3} = 183 \\ multiplying \: 2nd \: equation \: by \: 3 \\3a\sqrt{b}+3b\sqrt{a}=546 \: and \: \ adding \: we \: get \\ {( \sqrt{a}) }^{3 } + {( \sqrt{b} )}^{3} + 3 \sqrt{ab} ( \sqrt{a } + \sqrt{b} ) = 729 \\ {( \sqrt{a} + \sqrt{b} ) }^{3} = 729 \\ \sqrt{a} + \sqrt{b} = 9 \\ \\ from \: 2nd \: equation \\taking \:common\: \sqrt{ab} \:we \:get\\ \sqrt{ab}(\sqrt{a}+\sqrt{b}) \\ now \:putting \:value\: of\: \sqrt{a}+\sqrt{b} \\you\: will\:get\:\sqrt{ab} = \frac{182}{9} \\ now \\ {( \sqrt{a} + \sqrt{b}) }^{2} = a + b + 2 \sqrt{ab} \\ 81 - 2 \times \frac{182}{9} = a + b \\ a + b = \frac{365}{9}$

Answer 2

Given : a√a + b√b = 183   ----- (1)

Given : a√b + b√a = 182   ------ (2)

Let √a = x (or) a = x^2

Let √b = y (or) b = y^2.

Now, Equation (1) can be written as:

⇒ x^2(x) + y^2(y) = 183

⇒ x^3 + y^3 = 183     ------ (3)

Now, Equation (2) can be written as:

⇒ x^2(y) + y^2(x) = 182

⇒ x^2y + y^2x = 182      ------ (4)

On multiplying (3) by both sides, we get

⇒ 3(x^2y + y^2x) = 3 * 182

⇒ 3xy(x + y) = 546.     -----(5)

On adding (3) & (5), we get

⇒ x^3 + y^3 + 3xy(x + y) = 183 + 546

⇒ (x + y)^3 = 729

⇒ (x + y)^3 = (9)^3

⇒ x + y = 9      --------- (5)

Substitute (5) in (4), we get

⇒ x^2y + y^2x = 182

⇒ xy(x + y) = 182

⇒ xy = 182/9       ------ (6)

Now,

On squaring (5) on both sides, we get

⇒ (x + y)^2 = (9)^2

⇒ x^2 + y^2 + 2xy = 81

⇒ x^2 + y^2 + 2(182/9) = 81

⇒ x^2 + y^2 = 365/9.

Therefore, a + b = (365/9).

Hope it helps!