This question was uploaded on 01/01/22 on social media accounts.
Solution:
Let (a+b) = x and ab = y
We know that:
\[(a+b)^3 = a^3 + b^3 + 3ab(a+b)\]
\[\Rightarrow x^3 = a^3 + b^3 + 3xy\]
\[\Rightarrow a^3 + b^3 = x^3 - 3xy\]
Now given equation:
\[(a^2+1)(b^2+1)+25 = 10(a+b)\]
\[\Rightarrow a^2b^2+a^2+b^2+1+25 = 10(a+b)\]
\[\Rightarrow a^2b^2+a^2+\color{Red}{2ab}+b^2+26-\color{Red}{2ab} = 10(a+b)\]
\[\Rightarrow a^2b^2+(a+b)^2+26-2ab-10(a+b)=0\]
\[\Rightarrow y^2+x^2+26-2y-10x=0\]
\[\Rightarrow (x^2-10x+25)+(y^2-2y+1)=0\]
\[\Rightarrow (x-5)^2+(y-1)^2=0\]
Here both terms must be zero to satisfy the above equation:
\[\Rightarrow x-5=0 \Rightarrow x=5\]
and
\[\Rightarrow y-1=0 \Rightarrow y=1\]
Now we can find:
\[a^3+b^3=x^3-3xy=5^2-3(5)(1)=110\]
