Question

$\frac{a}{x}=\frac{b}{y}=\frac{c}{z}$ হলে প্রমাণ কর যে, $\frac{a^3+b^3+c^3}{x^3+y^3+z^3}=\frac{abc}{xyz}$

Answer 1

ধরি, $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=k$

$\therefore a=xk,b=yk,c=zk$

বামপক্ষ,
$=\frac{a^3+b^3+c^3}{x^3+y^3+z^3}$

$=\frac{\left(xk_{}\right)^3+\left(yk\right)^3+\left(zk\right)^3}{x^3+y^3+z^3}$

$=\frac{k^3\left(x^3+y^3+z^3\right)}{x^3+y^3+z^3}$

$=k^3$

ডানপক্ষ,
$=\frac{abc}{xyz}$

$=\frac{xk\cdot yk\cdot zk}{xyz}$

$=\frac{k^3xyz}{xyz}$

$=k^3$

$\therefore \frac{a^3+b^3+c^3}{x^3+y^3+z^3}=\frac{abc}{xyz}$ [proved]