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Algebra Level 3

2 + 5 3 + 2 − 5 3 = ? \large \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} = \, ?

Give your answer to 3 decimal places


The answer is 1.0000.

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Sharky Kesa by Sharky Kesa , 19, United Kingdom

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3 solutions

Let A = 2 + 5 3 \color{#E81990}{A}= \sqrt[3]{2+\sqrt{5}} , B = 2 − 5 3 \color{#EC7300}{B}=\sqrt[3]{2-\sqrt{5}} .
And, ⇒ S = A + B = 2 + 5 3 + 2 − 5 3 \Rightarrow\color{#D61F06}{S}=\color{#3D99F6}{A+B}=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}
Now, Cubing both sides.
⇒ S 3 = ( A + B ) 3 \Rightarrow \color{#D61F06}{S^3}=\color{#3D99F6}{(A+B)^3}
= A 3 + B 3 + 3 A B ( A + B ) =\color{#3D99F6}{A^3+B^3}+\color{#20A900}{3AB}\color{#3D99F6}{(A+B)}
= A 3 + B 3 + 3 A B S =\color{#3D99F6}{A^3+B^3}+\color{#20A900}{3AB}\color{#D61F06}{S} ...... ( ♧ ) (\color{#20A900}{♧}) [ Since , ( A + B ) = S \color{#3D99F6}{(A+B)}=\color{#D61F06}{S} ]
Thus,
⇒ A 3 + B 3 = ( 2 + 5 3 ) 3 + ( 2 − 5 3 ) 3 \Rightarrow \color{#3D99F6}{A^3+B^3}=(\sqrt[3]{2+\sqrt{5}})^{3}+(\sqrt[3]{2-\sqrt{5}})^{3}
= 2 + 5 + 2 − 5 =2+\sqrt{5}+2-\sqrt{5}
= 4 =\color{#302B94}{4}


⇒ 3 A B = 3 ( 2 + 5 3 ) ( 2 − 5 3 ) \Rightarrow \color{#20A900}{3AB}=3(\sqrt[3]{2+\sqrt{5}})(\sqrt[3]{2-\sqrt{5}})
= 3 ( 4 − 5 3 ) =3(\sqrt[3]{4-5})
= 3 ( − 1 3 ) =3(\sqrt[3]{-1})
= 3 × ( − 1 ) =3×(-1)
= − 3 =\color{#BA33D6}{-3}
Now, putting ( A 3 + B 3 ) = 4 \color{#3D99F6}{(A^3+B^3)}=\color{#302B94}{4} and ( 3 A B ) = − 3 \color{#20A900}{(3AB)}=\color{#BA33D6}{-3} in ( ♧ ) (\color{#20A900}{♧}) .
⇒ S 3 = 4 − 3 S \Rightarrow \color{#D61F06}{S^3}=\color{#302B94}{4}\color{#BA33D6}{-3}\color{#D61F06}{S}
S 3 + 3 S − 4 = 0 \color{#D61F06}{S^3}\color{#BA33D6}{+3}\color{#D61F06}{S}\color{#302B94}{-4}=0
( S − 1 ) ( S 2 + S + 4 ) = 0 (\color{#D61F06}{S}-1)(\color{#D61F06}{S^2}+\color{#D61F06}{S}\color{#302B94}{+4})=0
S − 1 = 0 \color{#D61F06}{S}-1=0
S = 1 \color{#D61F06}{S}=\color{#624F41}{\boxed{1}}

⇒ 2 + 5 3 + 2 − 5 3 = 1 . \Rightarrow \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}=\color{#624F41}{\boxed{1}}.

Note : There is a possibilty that ( S 2 + S + 4 = 0 ) (\color{#D61F06}{S^2}+\color{#D61F06}{S}\color{#302B94}{+4}=0) ,but this is not possible because its roots are non-real complex numbers.And the surds given is real.

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★【Nice coloring】★

Ashish Menon - 5 years ago

nice solution .. +1

Sabhrant Sachan - 5 years ago

Bro cong!!!!and ru solution and my solution is same but you know(-1)^(1÷3) is complex so who u write this -1tell me.....and no answer then ur solution were wrong..

VIneEt PaHurKar - 5 years ago

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Ask yourself what value should be multiplied 3 times to get a real one.
− 1 × − 1 × − 1 3 = − 1 \sqrt[3]{-1×-1×-1}=\boxed{-1} . However, there are three values where one is real and two are complex.
As you know ι = − 1 = ( − 1 ) 1 2 \iota=\sqrt{-1}=(-1)^{\frac{1}{2}} .

( − 1 ) 1 3 = ( − 1 ) 1 2 × 2 3 = ι 2 3 ≈ 0.500 + 0.86 ι (-1)^{\frac{1}{3}}=(-1)^{\frac{1}{2}×\frac{2}{3}}=\iota^{\frac{2}{3}}\approx0.500+0.86\iota

A Former Brilliant Member - 5 years ago

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Agreed. Well explained! & nice sol.+1 !

Rishabh Tiwari - 5 years ago

I did it the same way as Abhay....but how can you say that (-1)^(1/3) is complex.? In cube roots none of the values can be complex.!

Rishabh Tiwari - 5 years ago
Sharky Kesa
May 21, 2016

I have an algebraic-number-theoretic solution to this question. Let x = 2 + 5 3 + 2 − 5 3 x=\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} . Thus, we have

x − 2 + 5 3 − 2 − 5 3 = 0 x-\sqrt[3]{2+\sqrt{5}} - \sqrt[3]{2-\sqrt{5}} = 0

We have for numbers a a , b b and c c such that a + b + c = 0 a+b+c=0 , then a 3 + b 3 + c 3 = 3 a b c a^3+b^3+c^3=3abc . (This is not too hard to show and is left as an exercise for the reader.) Thus, we have

x 3 − ( 2 + 5 ) − ( 2 − 5 ) = 3 x ( 2 + 5 ) ( 2 − 5 ) 3 x 3 − 4 = 3 x − 1 3 x 3 + 3 x − 4 = 0 ( x − 1 ) ( x 2 + x + 4 ) = 0 \begin{aligned} x^3-(2+\sqrt{5})-(2-\sqrt{5})&=3x\sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})}\\ x^3-4 &= 3x\sqrt[3]{-1}\\ x^3+3x-4&=0\\ (x-1)(x^2+x+4)&=0 \end{aligned}

Note that x 2 + x + 4 x^2+x+4 has no real solutions, and we are assumed to take only the real cube root. Thus, the value must be 1. Therefore, 2 + 5 3 + 2 − 5 3 = 1 \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}=\boxed{1} .

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Interesting approach used here.

I'm not sure to what extent it is "number-theoretic". Seems mostly algebraic to me.

¶Superb¶ (+1)

Ashish Menon - 5 years ago

I have also used the same identity.

Aditya Sky - 5 years ago

Nicely done. +1 !

Rishabh Tiwari - 5 years ago

Same approach! A quite standard approach for finding values of this form.

展豪 張 - 5 years ago
Sam Bealing
May 22, 2016

2 + 5 3 = 1 + 5 2 \sqrt[3]{2+\sqrt{5}}=\dfrac{1+\sqrt{5}}{2}

2 − 5 3 = 1 − 5 2 \sqrt[3]{2-\sqrt{5}}=\dfrac{1-\sqrt{5}}{2}

2 + 5 3 + 2 − 5 3 = 1 − 5 2 + 1 + 5 2 = 1 \sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=\dfrac{1-\sqrt{5}}{2}+\dfrac{1+\sqrt{5}}{2}=\boxed{1}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

@Sam Bealing 2 + 5 3 = ϕ \sqrt[3]{2+\sqrt{5}}=\phi ,the golden ratio ,that's cool!Can you show how?

Rohit Udaiwal - 5 years ago

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⇒ 2 + 5 3 = 8 ( 2 + 5 ) 8 3 = 16 + 8 5 8 3 = 15 + 1 + 5 5 + 3 5 8 3 = ( 1 + 5 2 ) 3 3 = 1 + 5 2 = ϕ \Rightarrow \sqrt[3]{2+\sqrt{5}} =\sqrt[3]{\dfrac{8(2+\sqrt{5})}{8}}=\sqrt[3]{\dfrac{16+8\sqrt{5}}{8}}=\sqrt[3]{\dfrac{15+1+5\sqrt{5}+3\sqrt{5}}{8}}=\sqrt[3]{\left({\dfrac{1+\sqrt{5}}{2}}\right)^3}=\dfrac{1+\sqrt{5}}{2}=\phi

A Former Brilliant Member - 5 years ago

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Great job!That's correct :)

Rohit Udaiwal - 5 years ago

In the fourth step,

Shouldn't it be sq.root5 instead of sq.root3 ???

Rishabh Tiwari - 5 years ago

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@Rishabh Tiwari – Oh...Sorry fixed.

A Former Brilliant Member - 5 years ago

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@A Former Brilliant Member – PERFECT NOW!

Rishabh Tiwari - 5 years ago

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