If $a, b$ and $c$ are real numbers such that $a^2+b^2+c^2-a b-b c-a c \leq 0$, then $$ \left|\begin{array}{ccc} (a-b+1)^5 & b^7-c^7 & c^9-a^9 \ a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \ a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3)^1 \end{array} ight|= $$

Question

If $a, b$ and $c$ are real numbers such that $a^2+b^2+c^2-a b-b c-a c \leq 0$, then $$ \left|\begin{array}{ccc} (a-b+1)^5 & b^7-c^7 & c^9-a^9 \ a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \ a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3)^1 \end{array}ight|= $$

Aicharya · Accepted Answer

Given, $a^2+b^2+c^2-a b-b c-c a \leq 0$ $ 	herefore \frac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2ight] \leq 0 $ It is possible only $a=b=c$ $ \begin{aligned} & 	ext { Now, }\left|\begin{array}{ccc} (a-b+1)^5 & b^7-c^7 & c^9-a^9 \ a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \ a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3) 1 \end{array}ight| \ & \Rightarrow\left|\begin{array}{lll} 1 & 0 & 0 \ 0 & 8 & 0 \ 0 & 0 & 3 \end{array}ight|=24 \end{aligned} $

If $a, b$ and $c$ are real numbers such that $a^2+b^2+c^2-a b-b c-a c \leq 0$, then $$…

Solution