prantick wrote:answer 171

387 + 170 + 1 = 558

1 cube 1

2 cube 8

3 cube 27

4 cube 64

5 cube 125

6 cube 216

7 cube 343

8 cube 512

9 cube 729

729 - 558 = 171 (ans)

What have you done

Please read the question carefully.The question asked to find the three consecutive integers.

ahsaf wrote:i am confused

Here is the solution:

$1,170$ and $387$ are added with a integer.Suppose the integer is $a$.Let $a+1=x^{3},a+170=(x+1)^{3}$ and $a+387=(x+2)^{3}$

Now,$a+1=x^{3}\Rightarrow a=x^{3}-1.......(1)$.

$a+170=(x+1)^{3}\Rightarrow a=(x+1)^{3}-170.........(2)$.

$a+387=(x+2)^{3}\Rightarrow a=(x+2)^{3}-387...........(3)$.

From (1) and (2) we get,

$x^{3}-1=x^{3}+1+3x^{2}+3x-170\Rightarrow 3x^{2}+3x=168\Rightarrow 6x^{2}+6x=336..........(4)$.

From (3) we get,

$a+387=x^{3}+6x^{2}+12x+8\Rightarrow a=x^{3}+6x^{2}+12x-379$.

From (4) we get,$6x^{2}+6x=336$.

$\therefore a=x^{3}-43+6x$........(5)

From (1) and (5) we get,

$x^{3}-1=x^{3}-43+6x\Rightarrow x=7$.

$\therefore$ The digits are $7,8$ and $9$.