Question

Three numbers are in the ratio 1:2:3.The sum of their cubes is 7776 then the numbers are

Answer 1

Answer:   6 , 12 and 18

Step-by-step explanation:

given ,

numbers are in ratio , n1 :n2:n3 = 1:2:3

let n1 = 1x

    n2=2x

    n3=3x

accourding to qsn, (n1)^3+(n2)^3+(n3)^3= 7776

(1x)^3+(2x)^3+(3x)^3=7776

x^3+ 8x^3+27x^3=7776

36x^3= 7776

x^3= 216

x= 6

so , n1=6

n2= 2*6= 12

n3= 3*6 = 18

hope this will help......................

Answer 2

Step-by-step explanation:

SOLUTION

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.