If b is the mean proportional of a and c then ${(a - b)}^{3} : {(b - c)}^{3}$ equal

a^{\frac{3}{2}} : c^{\frac{3}{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

b^{2} : c^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a^{2} : c^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a^{3} : b^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $a^{\frac{3}{2}} : c^{\frac{3}{2}}$
Since b is the mean proportional of a and c, $b^{2} = a c, i . e ., b = \sqrt{a c} . ∴ \frac{{(a - b)}^{3}}{{(b - c)}^{3}} = \frac{{(a - \sqrt{a c})}^{3}}{{(\sqrt{a c} - c)}^{3}} = \frac{{[\sqrt{a} (\sqrt{a} - \sqrt{c})]}^{3}}{{[\sqrt{c} (\sqrt{a} - \sqrt{c})]}^{3}} = \frac{{(\sqrt{a})}^{3} {(\sqrt{a} - \sqrt{c})}^{3}}{{(\sqrt{c})}^{3} {(\sqrt{a} - \sqrt{c})}^{3}} = \frac{{(a 1 / 2)}^{3}}{{(c^{1 / 2})}^{3}} = \frac{a^{3 / 2}}{c^{3 / 2}}$

Suggest Corrections

Similar questions

a + b + c = 1; a^{2} + b^{2} + c^{2} = 2; a^{3} + b^{3} + c^{3} = 3

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

(c_{1}, c_{2}, c_{3})

x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0

\frac{(a^{2} - b^{2}) + {(b^{2} - c^{2})}^{3} + {(c^{2} - a^{2})}^{3}}{{(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3}}

Find $\frac{(a^{2} - b^{2})^{3} + (b^{2} - c^{2})^{3} + (c^{2} - a^{2})^{3}}{(a - b)^{3} + (b - c)^{3} + (c - a)^{3}} =$

\frac{(a^{2} - b^{2})^{3} + (b^{2} - c^{2})^{3} + (c^{2} - a^{2})^{3}}{(a - b)^{3} + (b - c)^{3} + (c - a)^{3}}

Join BYJU'S Learning Program