Question

The age of Mr. x last year was square of no.and it would be cube of next year.what is the least number of years he must wait for his age to becone the cube of nunber again?

Answer 1

let take initial age be x then after t years it would be x√3 th3n of take 2 yeqrs initially then cube is 8 then again cub3 means 2 the power 6 which is equals to 64 so he must wqit 64-8 =58 urs to be cibe again

Answer 2

Answer:

Mr. X must wait 18 years to make his age cube of his present age.

Solution:

Let the present age of Mr. X be a years.

Last year, Mr. X’s age will be $a^2$

Next year, his age will be $a^3$.

Let his present age be taken as 3.

Then,  

$\begin{aligned} a ^ { 2 } & = 3 ^ { 2 } = 9 \\\\ a ^ { 3 } & = 3 ^ { 3 } = 27 \end{aligned}$

Thus, number of years Mr. X must need to wait to become his age the cube of his present age will be,

$a ^ { 3 } - a ^ { 2 } = 27 - 9 = 18$

Hence, Mr. X must wait 18 years to make his age cube of his present age.