Question

যদি $a:b=b:c$ হয়, তবে প্রমাণ কর যে, $\frac{abc\left(a+b+c\right)^3}{\left(ab+bc+ca\right)^3}=1$

Answer 1

দেওয়া আছে,
$a:b=b:c$ বা, $\frac{a}{b}=\frac{b}{c}$ বা, $b^2=ac$

Left Hand Side,
$\frac{abc\left(a+b+c\right)^3}{\left(ab+bc+ca\right)^3}$

$=\frac{ac\cdot b\left(a+b+c\right)^3}{\left(ab+bc+ac\right)^3}$

[$\because b^2=ac$]

$=\frac{b^2\cdot b\left(a+b+c\right)^3}{\left(ab+bc+b^2\right)^3}$

$=\frac{b^3\left(a+b+c\right)^3}{\left\lbrace b\left(a+c+b\right)\right\rbrace^3}$

$=\frac{b^3\left(a+b+c\right)^3}{b^3\left(a+b+c\right)^3}$

$=1$

$=$ Right Hand Side

$\therefore$ Left Hand Side $=$ Right Hand Side

অর্থাৎ, $\frac{abc\left(a+b+c\right)^3}{\left(ab+bc+ca\right)^3}=1$ [Proved]