If a+b+c=11 and a b+b c+c a=20, then the value of the expression a³+b³+c³-3 a b c will be · ABC26GN1219

$(a+b+c)=11 \Rightarrow(a+b+c)^{2}=(11)^{2}=121$ $$\begin{aligned} & \Rightarrow a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=121 \\ & \Rightarrow a^{2}+b^{2}+c^{2}+2 \times 20=121 \\ & \Rightarrow a^{2}+b^{2}+c^{2}=121-40=81 . \end{aligned} \begin{aligned} \therefore \quad a^{3}+b^{3}+c^{3}-3 a b c & =(a+b+c) \\ & =\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\ & =11(81-20)=11 \times 61=671 . \end{aligned}$$