Given, a√a+b√b=32⋅⋅⋅⋅⋅⋅⋅(i) a√b+b√a=31⋅⋅⋅⋅⋅⋅⋅(ii) On squaring both sides Eqs. (i) and (ii), we get (a√a+b√b)2=322 [∵(a+b)2=a2+b2+2ab] ∴a3+b3+2ab√ab=1024⋅⋅⋅⋅⋅⋅⋅(iii) (a√b+b√a)2=312 a2b+b2a+2ab√ab=961⋅⋅⋅⋅⋅⋅⋅(iv) On subtracting Eq. (iv) from Eq. (iii), we get (a3+b3+2ab√ab)−(a2b+b2a+2ab√ab) =1024−961
a3+b3−a2b−b2a=63 a3−a2b−b2a+b3=63 a2(a−b)−b2(a−b)=63 (a2−b2)(a−b)=63 (a+b)(a−b)(a−b)=63 {∵a2−b2=(a+b)(a−b)} (a+b)(a−b)2=63 (a+b) and (a−b)2 must be a co-prime number, by factorising 63 we get 7 and 9 that can satisfy the above equation. ∵a+b=7 (a−b)2=9⇒a−b=±3 If we take, a−b=−3, we will get b negative that can not be possible, ∵a+b=7 a−b=3 a=5 ⇒b=2 Value of