Question

three numbers are in the ratio 1:2:3 The sum of their cubes is 79092 find ​

Answer 1

Answer :–

• The first number = 13.
• The second number = 26.
• The third number = 39.

Given :–

• Three numbers are in the ratio = 1 : 2 : 3.
• The sum of their cubes = 79092.

To Find :–

• Find all the three numbers.

Solution :–

Let,

The first number be 1x.

The second number be 2x.

The third number be 3x.

According to the question,

The sum of their cubes is 79092.

Thay means,

$\rightarrow {(x)}^{3} + {(2x)}^{3} + {(3x)}^{3} = 79092$

Now, open the brackets.

$\rightarrow {x}^{3} + {8x}^{3} + {27x}^{3} = 79092$

$\rightarrow{36x}^{3} = 79092$

$\rightarrow {x}^{3} = \cancel\dfrac{79092}{36}$

$\rightarrow{x}^{3} = \cancel \dfrac{39546}{18}$

$\rightarrow {x}^{3} = \cancel \dfrac{19773}{9}$

$\rightarrow {x}^{3} = \cancel\dfrac{6591}{3}$

$\rightarrow {x}^{3} = 2197$

$\rightarrow x = \sqrt[3]{2197}$

$\rightarrow x = 13$

So,

we get the value of x is 13.

Now, we have to find the numbers.

So,

The first number = 1x = 1 × 13 = 13.

The second number = 2x = 2 × 13 = 26.

The third number = 3x = 3 × 13 = 39.

Hence,

The three numbers are 13, 36 and 39.

Verification :–

According to the question,

The sum of their cubes is 79092.

We have,

• First number = 13.
• Second number = 26.
• Third number = 39.

Now, substitute the values of all the three numbers.

$\rightarrow {(13)}^{3} + {(26)}^{3} + {(39)}^{3} = 79092$

$\rightarrow 2197 + 17576+ 59319 = 79092$

$\rightarrow 79092 = 79092$

Hence Verified !!

Answer 2

Step-by-step explanation:

Answer :–

The first number = 13.

The second number = 26.

The third number = 39.

Given :–

Three numbers are in the ratio = 1 : 2 : 3.

The sum of their cubes = 79092.

To Find :–

Find all the three numbers.

Solution :–

Let,

The first number be 1x.

The second number be 2x.

The third number be 3x.

According to the question,

The sum of their cubes is 79092.

Thay means,

\rightarrow {(x)}^{3} + {(2x)}^{3} + {(3x)}^{3} = 79092→(x)

3

+(2x)

3

+(3x)

3

=79092

Now, open the brackets.

\rightarrow {x}^{3} + {8x}^{3} + {27x}^{3} = 79092→x

3

+8x

3

+27x

3

=79092

\rightarrow{36x}^{3} = 79092→36x

3

=79092

\rightarrow {x}^{3} = \cancel\dfrac{79092}{36}→x

3

=

36

79092

\rightarrow{x}^{3} = \cancel \dfrac{39546}{18}→x

3

=

18

39546

\rightarrow {x}^{3} = \cancel \dfrac{19773}{9}→x

3

=

9

19773

\rightarrow {x}^{3} = \cancel\dfrac{6591}{3}→x

3

=

3

6591

\rightarrow {x}^{3} = 2197→x

3

=2197

\rightarrow x = \sqrt[3]{2197}→x=

3

2197

\rightarrow x = 13→x=13

So,

we get the value of x is 13.

Now, we have to find the numbers.

So,

The first number = 1x = 1 × 13 = 13.

The second number = 2x = 2 × 13 = 26.

The third number = 3x = 3 × 13 = 39.

Hence,

The three numbers are 13, 36 and 39.

Verification :–

According to the question,

The sum of their cubes is 79092.

We have,

First number = 13.

Second number = 26.

Third number = 39.

Now, substitute the values of all the three numbers.

\rightarrow {(13)}^{3} + {(26)}^{3} + {(39)}^{3} = 79092→(13)

3

+(26)

3

+(39)

3

=79092

\rightarrow 2197 + 17576+ 59319 = 79092→2197+17576+59319=79092

\rightarrow 79092 = 79092→79092=79092

Hence Verified !!