Question

if any have potential so do this question its my challenge that no can do it a√a+b√b=183 and a√b+b√a=182 then find 9/5(a+b)

Answer 1

Let:

$\sqrt{a} = x$

= > a = x^2 ------ (1)


Let 

$\sqrt{b} = y$

= > b = y^2  ------ (2)

Now,

Given Equation becomes:

= > x^3 + y^3 = 183 ----- (3)

= > x^2y + y^2x = 182  ---- (4)

On solving (1) & (2) * 3, we get

= > x^3 + y^3 + 3(x^2y + y^2x) = 183 + 3(182)

= > x^3 + y^3 + 3x^2y + 3y^2x = 183 + 546

= > x^3 + y^3 + 3x^2y + 3y^2x = 729

It is in the form of a^3 + b^3 + 3a^2b + 3ab^2 = (a + b)^3

= > (x + y)^3 = 729

= > (x + y) = 9  ---------- (5)

We know that:

a^3 + b^3 = (a+ b)(a^2 - ab + b^2).

Likewise,

x^3 + y^3 :

= >  (x + y)(x^2 + y^2 - xy) = 183

= >  (x + y)(x^2 + y^2) - xy(x + y) = 183

= > (9)(x^2 + y^2) - 182 = 183

= > (9)(x^2 + y^2) = 183 + 182

= > (9)(x^2 + y^2)  = 365

= > (x^2 + y^2) = 365/9  ------- (6)

From (1) &(2), Equation (6) can be written as:

= > a  + b = 365/9  -------- (7)


Now,

Given Equation is 9/5(a + b)

$= \ \textgreater \ \frac{9}{5} * \frac{365}{9}$

$= \ \textgreater \ \frac{365}{9}$

$= \ \textgreater \ 73$




Therefore, the value of 9/5(a + b) = 73.



Hope this helps!