Question

If sinA+sinB+sinC=0, show that sin3A+sin3B+sin3B+sin3C+12sinA sinB sinC=O

Answer 1

$\textbf{Concept:}$

$\text{If a+b=c=0 then }a^3+b^3+c^3=3abc$

$\textbf{Given:}$

$\text{sinA+sinB+sinC=0}$

$\text{Using the given concept, we get}$

$sin^3A+sin^3B+sin^3C=3\;sinA\;sinB\;sinC$

$\text{using}$

$\bf\;sin3A=3\;sinA-4\;sin^3A\;\implies\;sin^3A=\frac{1}{4}[3\;sinA-sin3A]$

$\frac{1}{4}[3\;sinA-sin3A]+\frac{1}{4}[3\;sinB-sin3B]+\frac{1}{4}[3\;sinC-sin3C]=3\;sinA\;sinB\;sinC$

$3\;sinA-sin3A+3\;sinB-sin3B+3\;sinC-sin3C=12\;sinA\;sinB\;sinC$

$3(sinA+sinB+sinC)-sin3A-sin3B-sin3C=12\;sinA\;sinB\;sinC$

$3(0)-sin3A-sin3B-sin3C=12\;sinA\;sinB\;sinC$

$-sin3A-sin3B-sin3C-12\;sinA\;sinB\;sinC=0$

$sin3A+sin3B+sin3C+12\;sinA\;sinB\;sinC=0$

$\therefore\boxed{\bf\;sin3A+sin3B+sin3C+12\;sinA\;sinB\;sinC=0}$

Answer 2

Answer:

-12sinAsinBsinC

Step-by-step explanation:

Refer to the attachment given below.