Question

The difference of the cubes of two positive numbers is 2402 and the cube of their difference is 8. Find the smaller number.

Answer 1

ANSWER:

Given:

• Difference of cubes of 2 numbers = 2402
• Cube of difference of the numbers = 8

To Find:

• The smaller number

Solution:

Let's assume that, the bigger number be x and smaller be y.

We are given that,

⇒ Difference of cubes of 2 numbers = 2402

So,

$\implies x^3-y^3=2402$

We are also given that,

⇒ Cube of difference of the numbers = 8

So,

$\implies(x-y)^3=8$

Taking cube-root on both sides,

$\implies\sqrt[3]{(x-y)^3}=\sqrt[3]{8}$

So,

$\implies x-y=2$

We know that

$\hookrightarrow a^3-b^3=(a-b)(a^2+b^2+ab)$

And,

$\hookrightarrow a^2+b^2=(a-b)^2+2ab$

Hence,

$\hookrightarrow a^3-b^3=(a-b)((a-b)^2+2ab+ab)$

$\hookrightarrow a^3-b^3=(a-b)((a-b)^2+3ab)$

Here, a = x and b = y. So,

$\implies x^3-y^3=2402$

$\implies (x-y)((x-y)^2+3xy)=2402$

Substituting the (x - y) with 2,

$\implies (2)((2)^2+3xy)=2402$

$\implies 2(4+3xy)=2402$

Transposing 2 to RHS,

$\implies (4+3xy)=\dfrac{2402}{2}$

$\implies 4+3xy=1201$

Transposing 4 to RHS,

$\implies 3xy=1201-4$

$\implies 3xy=1197$

Transposing 3 to RHS,

$\implies xy=\dfrac{1197}{3}$

$\implies xy=399$

So,

$\implies x=\dfrac{399}{y}$

We had,

$\implies x-y=2$

Hence,

$\implies \dfrac{399}{y}-y=2$

Taking LCM,

$\implies\dfrac{399-y^2}{y}=2$

Transposing y to RHS,

$\implies399-y^2=2y$

So,

$\implies y^2+2y-399=0$

Splitting the middle term,

$\implies y^2+21y-19y-399=0$

$\implies y(y+21)-19(y+21)=0$

$\implies (y+21)(y-19)=0$

Hence,

$\implies y=-21\:\:\&\:\:y=19$

We are given that, the numbers are positive. So,

$\implies\bf y=19$

Hence, the smaller number is 19.