Let a,b,cR be all non-zero and satisfy a^3+b^3+c^3=2.

Question

Let $a,b,c\in\mathbb{R}$ be all non-zero and satisfy $a^3+b^3+c^3=2$. If the matrix $A=\begin{pmatrix}a&b&c\b&c&a\c&a&b\end{pmatrix}$ satisfies $A^TA=I$, then a value of $abc$ can be:

Accepted Answer

$A^TA=I$ gives $\sum a^2=1$ and $\sum ab=0$ (sum of pairwise products of $a,b,c$).

Using the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=(a+b+c)(1-0)=(a+b+c)$.

Also $(a+b+c)^2=\sum a^2+2\sum ab=1+0=1\Rightarrow a+b+c=\pm1$.

So $2-3abc=\pm1$. Taking $2-3abc=1$: $abc=\dfrac13$.

Solutions · 1