In a , prove that ^3A (B-C)+ ^2B (C-A)+ ^3C (A-B) =3

Question

In a $\triangle\text{ABC},$ prove that $\sin^3\text{A}\cos(\text{B}-\text{C})+\sin^2\text{B}\cos(\text{C}-\text{A})+\sin^3\text{C}\cos(\text{A}-\text{B})\\=3\sin\text{A}\sin\text{B}\sin\text{C}$

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Answer

Let $\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}=\text{k}...(1)$ $\text{LHS}=\sin^3\text{A}\cos(\text{B}-\text{C})+\sin^3\text{B}\cos(\text{C}-\text{A})+\sin^3\text{C}\cos(\text{A}-\text{B})$ $=\sin^2\text{A}\{\sin\text{A}\cos(\text{B}-\text{C})\}+\sin^2\text{B}\{\sin\text{B}\cos(\text{C}-\text{A})\}\\ \ \ \ \ \ \ \ +\sin^2\text{A}\{\sin\text{A}\cos(\text{A}-\text{B})\}$ $=\frac{\text{a}^2}{\text{k}^2}\{\sin\text{A}\cos(\text{B}-\text{C})\}+\frac{\text{b}^2}{\text{k}^2}\{\sin\text{B}\cos(\text{C}-\text{A})\}+\frac{\text{c}^2}{\text{k}^2}\{\sin\text{A}\cos(\text{A}-\text{B})\}$ $=\frac{1}{2\text{k}^2}\big[\text{a}^2\{2\sin\text{A}\cos(\text{B}-\text{C})\}+\text{b}^2\{2\sin\text{B}\cos(\text{C}-\text{A})\}\\ \ \ \ \ \ \ \ \ \ +\text{c}^2\{2\sin\text{A}\cos(\text{A}-\text{B})\}\big]$ $=\frac{1}{2\text{k}^2}\big[\text{a}^2\{2\sin(\pi-(\text{B + C}))\cos(\text{B} - \text{C})\}+\text{b}^2\{2\sin(\pi-(\text{A + C}))\cos(\text{C}-\text{A})\}\\ \ \ \ \ \ \ \ \ \ +\text{c}^2\{2\sin(\pi-(\text{B + C}))\cos(\text{A}-\text{B})\}\big]$ $=\frac{1}{2\text{k}^2}\big[\text{a}^2\{2\sin(\text{B + C})\cos(\text{B}-\text{C})\}+\text{b}^2\{2\sin(\text{C + A})\cos(\text{C}-\text{A})\}\\ \ \ \ \ \ \ \ \ \ \ +\text{c}^2\{2\sin(\text{A + B})\cos(\text{A}-\text{B})\}\big]$ $=\frac{1}{2\text{k}^2}\big[\text{a}^2\{\sin2\text{B}+\sin2\text{C}\}+\text{b}^2\{\sin2\text{C}+\sin2\text{A}\}+\text{c}^2\{\sin2\text{A}+\sin2\text{B}\}\big]$ $=\frac{1}{2\text{k}^2}\big[2\text{a}^2\{\sin\text{B}\cos\text{B}+\sin\text{C}\cos\text{C}\}+2\text{b}^2\{\sin\text{C}\cos\text{C}+\sin\text{A}\cos\text{A}\}\\ \ \ \ \ \ \ \ \ \ +2\text{c}^2\{\sin\text{A}\cos\text{A}+\sin\text{B}\cos\text{B}\}\big]$ $=\frac{1}{2\text{k}^3}\big[2\text{a}^2\{\text{k}\sin\text{B}\cos\text{B}+\text{k}\sin\text{C}\cos\text{C}\}+2\text{b}^2\{\text{k}\sin\text{C}\cos\text{C}+\text{k}\sin\text{A}\cos\text{A}\}\\ \ \ \ \ \ \ \ \ \ +2\text{c}^2\{\text{k}\sin\text{A}\cos\text{A}+\text{k}\sin\text{B}\cos\text{B}\}\big]$ $=\frac{1}{\text{k}^3}\big[\text{a}^2\{\text{b}\cos\text{B + c}\cos\text{C}\}+2\text{b}^2\{\text{c}\cos\text{C + a}\cos\text{A}\}\\ \ \ \ \ \ \ \ +2\text{c}^2\{\text{a}\cos\text{A + a}\cos\text{B}\}\big]$ $=\frac{1}{\text{k}^3}\big[\text{ab(a}\cos\text{B + b}\cos\text{A})+\text{bc(b}\cos\text{C + c}\cos\text{B})+\text{ac(a}\cos\text{ C + c}\cos\text{A})\big]$ $=\frac{1}{\text{k}^3}(\text{abc + bca + acb})$ $=3\text{abc}\times\frac{1}{\text{k}^3}$ $=3\sin\text{A}\sin\text{B}\sin\text{C}\times\frac{1}{\text{k}^3}\times\text{k}^3$ $=3\sin\text{A}\sin\text{B}\sin\text{C}$ $=\text{RHS}$ Hence proved.