<s>[INST] If in a triangle $ABC$,$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A$,then prove that triangle will be equilateral triangle If in a triangle $ABC$ we have
$$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A\ ,$$
then the triangle will be equilateral triangle.
I tried but except few steps,could not prove it.  Putting $A=B=C=\frac{\pi}{3}$ makes both sides equal.  How should i prove it?Please help me.
 [/INST] As $A+B+C=\pi,\cos(A+B)=\cdots=-\cos C$ 
and using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$2\cos A\cos B\cos C=\cos C[\cos(A-B)+\cos(A+B)]$$
$$=-\cos(A+B)\cos(A-B)-\cos^2C$$
$$=-(\cos^2A-\sin^2B)-\cos^2C$$
$$=1-(\cos^2A+\cos^2B+\cos^2C)$$
So, the equation reduces to $$\sum(\cos A-\cos B)^2=0$$
Can you take it from here?
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