Prove that: $a^3\cos{B}\cos{C} + b^3\cos{C}\cos{A} + c^3\cos{A}\cos{B} = abc(1-2\cos{A}\cos{B}\cos{C}).$

Question

In $\Delta ABC$, prove that: $$a^3\cos{B}\cos{C} + b^3\cos{C}\cos{A} + c^3\cos{A}\cos{B} = abc(1-2\cos{A}\cos{B}\cos{C}).$$

I know that $$a^3\cos{(B-C)}+b^3\cos{(C-A)}+ c^3\cos{(A-B)}=3abc$$ and $$\cos^2{A}+\cos^2{B}+\cos^2{C} + 2\cos{A}\cos{B}\cos{C}=1.$$ Can these two results be used to prove the statement above? Any kind of help/suggestions will be highly appreciated.

Answer 1

Hints: By sine-law, $a=2R\sin A$, $b=2R\sin B$, and $c=2R\sin C$, where $R$ is the radius of the circumcircle of the triangle $\Delta ABC$. Writing $a,b,c$ in terms of $\sin A,\sin B,\sin C$ and cancelling the $R^{3}$ on both sides, the expression reduces to another expression involving $\sin A,\sin B,\sin C,\cos A,\cos B,\cos C$ only. Further write $C=\pi-A-B$, we can reduce the expression so that it contains $\sin A,\sin B,\cos A,\cos B$ only. Now, $A$ and $B$ are essentially not restricted. Therefore, we can anticipate that the expression arises from some trigonometric identities involving two variables.

Answer 2

$$a^3\cos{(B-C)}+b^3\cos{(C-A)}+ c^3\cos{(A-B)}=3abc$$ gives $$\sum_{cyc}a^2\frac{\cos B}b\frac{\cos C}c+\sum_{cyc}a^2\frac{\sin B}b\frac{\sin C}c=3.$$ By law of sines, i.e., $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$, we have $$\sum_{cyc}a^2\frac{\cos B}b\frac{\cos C}c+\sum_{cyc}a^2\frac{\sin^2 A}{a^2}=3,$$ i.e. $$\sum_{cyc}a^2\frac{\cos B}b\frac{\cos C}c+\sum_{cyc}(1-\cos^2 A)=3$$ $$\sum_{cyc}a^2\frac{\cos B}b\frac{\cos C}c-\sum_{cyc}\cos^2 A=0$$ $$\sum_{cyc}a^3\cos B\cos C=abc\sum_{cyc}\cos^2 A$$ and the title identity follows now by using the second identity you gave: $$\cos^2{A}+\cos^2{B}+\cos^2{C} + 2\cos{A}\cos{B}\cos{C}=1.$$

Answer 3

$$2a^3\cos B\cos C=a^3(\cos(B-C)+\cos(B+C))=a^3\cos(B-C)-a^3\cos A$$

$$a^3\cos A$$ $$=(2R)^3\sin^3A\cos A$$ $$=2R^3(2\sin^2A)\sin2A$$ $$=2R^3(1-\cos2A)\sin2A$$ $$=2R^3(\sin2A)-R^3(\sin4A)$$ Use Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle for $\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C$

Use this for $\sin4A+\sin4b+\sin4C=\cdots=-4(8\sin A\sin B\sin C)\cos A\cos B\cos C$

Can you take it from here?