If $$a,b\in R$$ such that $$a\sqrt{a} +b\sqrt{b} =183$$ and $$b\sqrt{a} +a\sqrt{b} =182$$, find the value of $$\dfrac{9}{5}(a+b) $$

Question

If $$a,b\in R$$ such that $$a\sqrt{a}  +b\sqrt{b}  =183$$ and $$b\sqrt{a}  +a\sqrt{b}  =182$$, find the value of $$\dfrac{9}{5}(a+b)  $$

Uniteasy Admin · Accepted Answer

Given conditions,
$$a\sqrt{a} +b\sqrt{b} =183$$n
$$b\sqrt{a} +a\sqrt{b} =182$$n
Let
$$x=\sqrt{a}, y=\sqrt{b}  $$
That is
$$a=x^2, b=y^2$$
Given conditions (1) and (2) are converted to
$$x^3+y^3 = 183$$n
$$xy^2+x^2y = 182$$n
Apply the formula for the cube of a binomial,
$$(x+y)^3 = x^3+3x^2y+3xy^2+y^3$$
$$=(x^3+y^3)+3(xy^2+x^2y)$$
$$=183+3	imes 182 $$
$$=729=9^3$$
Obtain the value of $$x+y $$ by taking the cube root of both sides of the equation:
$$x+y = 9$$n
Obtain the value of $$xy$$ by factoring out $$xy$$ from equation (4) 
$$xy(x+y) = 182 $$
$$xy = \dfrac{182}{x+y} $$
$$=\dfrac{182}{9} $$n
Square both sides of equation (5) 
$$(x+y)^2 = 81$$
$$x^2+y^2+2xy = 81$$
Substitute the value of $$xy$$
$$x^2+y^2 = 81-2\cdotp \dfrac{182}{9}  =\dfrac{365}{9} $$
Therefore,
$$a+b = \dfrac{365}{9} $$
Now we can get the final value
$$\dfrac{9}{5}(a+b) =\dfrac{9}{5}\cdotp \dfrac{365}{9} =73  $$

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