Find-the-cube-root-of-55-63-2-

Question Number 17018 by tawa tawa last updated on 29/Jun/17

Find the cube root of : 55 + 63 \sqrt{2}

Commented by prakash jain last updated on 29/Jun/17

{(a + b \sqrt{2})}^{3} = 55 + 63 \sqrt{2}

a^{3} + 3 \sqrt{2} a^{2} b + 6 b^{2} a + 2 \sqrt{2} b^{3} = 55 + 63 \sqrt{2}

a^{3} + 6 b^{2} a = 55 (i)

3 a^{2} b + 2 b^{3} = 63 (i i)

2 b^{3} = 3 (21 - a^{2} b)

b i n t e g e r i m p l i e s b = 3

9 a^{2} + 54 = 63

a = 1

c h e c k i n (i)

1 + 54 = 55 \dots (c o r r e c t)

\sqrt[3]{55 + 63 \sqrt{2}} = 1 + 3 \sqrt{2}

Commented by tawa tawa last updated on 29/Jun/17

God bless you sir .

Commented by mrW1 last updated on 30/Jun/17

can generally be said ?

{(a + b \sqrt{2})}^{n} = c + d \sqrt{2}

{(a + b \sqrt{3})}^{n} = c + d \sqrt{3}

\dots \dots

Commented by prakash jain last updated on 30/Jun/17

Not always .

Commented by prakash jain last updated on 30/Jun/17

if we change the question to

55 + 62 \sqrt{2} the answer wont be

of the form a + b \sqrt{2} .

i just tried this approach just

to see if a simple answer is possible .

Commented by mrW1 last updated on 30/Jun/17

is {(a + b \sqrt{2})}^{n} not always of the form

c + d \sqrt{2} ?

{(a + b \sqrt{2})}^{n} = \underset{k = 0}{\sum^{n}} C_{n}^{k} a^{n - k} {(b \sqrt{2})}^{k}

= C_{n}^{0} a^{n} + C_{n}^{2} a^{n - 2} b^{2} {(\sqrt{2})}^{2} + C_{n}^{4} a^{n - 4} b^{4} {(\sqrt{2})}^{4} + \dots

+ C_{n}^{1} a^{n - 1} b (\sqrt{2}) + C_{n}^{3} a^{n - 3} b^{3} {(\sqrt{2})}^{3} + C_{n}^{5} a^{n - 5} b^{5} {(\sqrt{2})}^{5} + \dots

= C_{n}^{0} a^{n} + {2 C}_{n}^{2} a^{n - 2} b^{2} + 2^{2} C_{n}^{4} a^{n - 4} b^{4} + \dots

+ C_{n}^{1} a^{n - 1} b \sqrt{2} + {2 C}_{n}^{3} a^{n - 3} b^{3} \sqrt{2} + 2^{2} C_{n}^{5} a^{n - 5} b^{5} \sqrt{2} + \dots

= [C_{n}^{0} a^{n} + {2 C}_{n}^{2} a^{n - 2} b^{2} + 2^{2} C_{n}^{4} a^{n - 4} b^{4} + \dots]

+ [C_{n}^{1} a^{n - 1} b + {2 C}_{n}^{3} a^{n - 3} b^{3} + 2^{2} C_{n}^{5} a^{n - 5} b^{5} + \dots] \sqrt{2}

= c + d \sqrt{2}

a, b, c, d = integer

Commented by mrW1 last updated on 30/Jun/17

but {(c + d \sqrt{2})}^{\frac{1}{n}} \neq a + b \sqrt{2} ? ? ?

Commented by tawa tawa last updated on 30/Jun/17

Thanks sirs . God bless you .