Question

three number are in the ratio of 1 ratio 2 ratio 3 ratio the sum of their cube is 7776 find the number

Answer 1

Sol : Let x is included in numbers.

So, numbers = x , 2 x and 3 x.

It is given that sum of their cubes is equal to 7,776.

So,

 ( x )³ + ( 2 x )³ + ( 3 x )³ = 7,776.

 x³ + 8 x³ + 27 x³ =7,776

 36 x³ = 7,776

     x³ =7,776 / 36

     x³ = 216

    x =³√216

   x = 6.

Numbers = x = 6 , 2 x = 2 * 6 = 12 and 3 x = 3 * 6 = 18.

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Answer 2

Step-by-step explanation:

SOLUTION

GIVEN,

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

The sum of their cubes is 7776.

FIND,

The numbers.

SO,

Let the three numbers be x, 2x and 3x.

Then,

${x}^{3} + (2x)^{3} + (3x)^{3} = 7776$

$= > {x}^{3} + {8x}^{3} + {27x}^{3} = 7776$

$= > {36x}^{3} = 7776$

$= > x^{3} = \frac{7776}{36}$

$= > {x}^{3} = 216$

$= > x = \sqrt[3]{216}$

$= > x = \sqrt[3]{2 \times 2 \times 2 \times 3 \times 3 \times 3}$

$= > x = 2 \times 3$

$= > x = 6$

Therefore,

x = 6, 2x = 2×6 = 12 and 3x = 3×6 = 18.

Hence,

the three numbers are 6, 12, 18.