Question

The cube root of the sum of three consecutive odd natural number is equal to the square root of cube root of 225. Find the sum of cubes of the number.

Answer 1

Answer:

$\red {Sum \: of \: cubes \: of \: the \: numbers}$

$\green {= 495}$

Step-by-step explanation:

$Let \: x , (x+2) \: and \: (x+4) \: are \: three \\consecutive \: odd \: natural \: numbers$

/* According to the problem given

$\sqrt[3]{x+(x+2)+(x+4)} = \sqrt{ \sqrt[3]{225}}$

$\implies \sqrt[3]{3x+6} = \sqrt[3]{\sqrt{15^{2}}}$

$\implies \sqrt[3]{3x+6} = \sqrt[3]{15}$

$\implies 3x+6 = 15$

$\implies 3x = 15 - 6$

$\implies 3x = 9$

$\implies x = \frac{9}{3} = 3$

$Three \: consecutive \: odd \: natural \\numbers \: are \: 3,5 \:and \: 7$

$Sum \: of \: cubes \: of \: the \: numbers \\= 3^{3} + 5^{3} + 7^{3} \\= 27 + 125 + 343\\= 495$

Therefore.,

$\red {Sum \: of \: cubes \: of \: the \: numbers}$

$\green {= 495}$

•••♪

Answer 2

Answer:

495

Step-by-step explanation:

according it question

let we take x, [x + 2], [x + 4]

= $\sqrt[3]{x + [x + 2] + [x + 4]}$ = $\sqrt[3]{\sqrt 225}$

= $\sqrt[3]{3x + 6}$ = $\sqrt[3]{\sqrt 15^{2} }$

= $\sqrt[3]{3x + 6\\}$ =$\sqrt[3]{15}$

= 3x + 6 = 15

= 3x = 15 -- 6

= 3x = 9

= x = 9/3

= x = 3 ans

therefore $3^{3} + 5^{3} + 7^{3}$

27 + 125 + 343 = 495 ans