Cube roots escape

Algebra Level 5

5 3 − 4 3 × 3 = a 3 + b 3 − c 3 , \sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{a} + \sqrt[3]{b} - \sqrt[3]{c},

where a , b a, b and c c are positive integers. What is the value of a + b + c a+b+c ?


The answer is 47.

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Calvin Lin by Calvin Lin

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3 solutions

George G
Dec 21, 2013

Let x , y , v x, y, v be positive real such that x 3 = 5 , y 3 = 4 , v 3 = 2 x^3=5, y^3=4, v^3=2 . So y = v 2 , y 2 = 2 v , v y = 2 y=v^2, y^2=2v, vy=2 . ( x − y ) ( x + y ) 2 = x 3 − y 3 + x 2 y − x y 2 = 1 + ( x v ) 2 − 2 x v = ( x v − 1 ) 2 (x-y)(x+y)^2 = x^3-y^3 + x^2y-xy^2= 1 + (xv)^2 - 2xv = (xv-1)^2 Hence 9 ( x − y ) = ( x − y ) ( x 3 + y 3 ) = ( x − y ) ( x + y ) 2 ( x 2 − x y + y 2 ) = ( x v − 1 ) ( x 2 − x y + y 2 ) = x 3 v − x 2 v y + x v y 2 − x 2 + x y − y 2 = 3 v + 3 x y − 3 x 2 = 3 ( v + x y − x 2 ) . 9\sqrt{(x-y)} = \sqrt{(x-y)} (x^3+y^3)=\sqrt{(x-y)(x+y)^2}(x^2-xy+y^2) \\ = (xv-1)(x^2-xy+y^2) = x^3v-x^2vy+xvy^2 - x^2+xy-y^2\\ =3v +3xy- 3x^2 = 3(v +xy- x^2). That is 5 3 − 4 3 × 3 = 2 3 + 20 3 − 25 3 \sqrt{\sqrt[3]{5} -\sqrt[3]{4} }\times{3} = \sqrt[3]{2} + \sqrt[3]{20} - \sqrt[3]{25}

17 Helpful 0 Interesting 0 Brilliant 0 Confused

Sorry for adding my solution here. I am amazed by your solution but I think something can be improved there.

Similarly, let

x 3 = 5 x^3=5

y 3 = 4 y^3=4

9 \strut x − y 9\sqrt{\strut x-y}

= ( x 3 + y 3 ) \strut x − y =(x^3+y^3)\sqrt{\strut x-y}

= ( x 2 − x y + y 2 ) \strut ( x − y ) ( x + y ) 2 =(x^2-xy+y^2)\sqrt{\strut(x-y)(x+y)^2}

= ( x 2 − x y + y 2 ) \strut x 3 + x 2 y − x y 2 − y 3 =(x^2-xy+y^2)\sqrt{\strut x^3+x^2y-xy^2-y^3}

Now we can substitute the value already

= ( \strut 25 3 − \strut 20 3 + \strut 16 3 ) \strut 5 + \strut 100 3 − \strut 80 3 − 4 =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut 5+\sqrt[3]{\strut 100}-\sqrt[3]{\strut80}-4}

= ( \strut 25 3 − \strut 20 3 + \strut 16 3 ) \strut \strut 10 3 2 − 2 \strut 10 3 + 1 =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut \sqrt[3]{\strut 10}^2-2\sqrt[3]{\strut10}+1}

= ( \strut 25 3 − \strut 20 3 + \strut 16 3 ) \strut ( \strut 10 3 − 1 ) 2 =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut (\sqrt[3]{\strut 10}-1)^2}

= ( \strut 25 3 − \strut 20 3 + \strut 16 3 ) ( \strut 10 3 − 1 ) =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})(\sqrt[3]{\strut 10}-1)

= \strut 250 3 − \strut 200 3 + \strut 160 3 − \strut 25 3 + \strut 20 3 − \strut 16 3 =\sqrt[3]{\strut 250}-\sqrt[3]{\strut 200}+\sqrt[3]{\strut 160}-\sqrt[3]{\strut 25}+\sqrt[3]{\strut 20}-\sqrt[3]{\strut 16}

= 5 \strut 2 3 − 2 \strut 25 3 + 2 \strut 20 3 − \strut 25 3 + \strut 20 3 − 2 \strut 2 3 =5\sqrt[3]{\strut 2}-2\sqrt[3]{\strut 25}+2\sqrt[3]{\strut 20}-\sqrt[3]{\strut 25}+\sqrt[3]{\strut 20}-2\sqrt[3]{\strut 2}

= 3 \strut 2 3 − 3 \strut 25 3 + 3 \strut 20 3 =3\sqrt[3]{\strut 2}-3\sqrt[3]{\strut 25}+3\sqrt[3]{\strut 20}

Hence

3 \strut \strut 5 3 − \strut 4 3 3\sqrt{\strut \sqrt[3]{\strut 5}-\sqrt[3]{\strut 4}}

= \strut 2 3 − \strut 25 3 + \strut 20 3 =\sqrt[3]{\strut 2}-\sqrt[3]{\strut 25}+\sqrt[3]{\strut 20}

Christopher Boo - 7 years, 2 months ago

Very elegant! Can you share an insight on how you thought of that?

Alexander Borisov - 7 years, 5 months ago

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The letters x , y , v x,y,v are just convenient for me to type in Latex. It's a bunch of numbers with cubic roots on my scratch paper. I am not sure I have any insight to share. It's mainly by trial and error. First of all, I suspected that the RHS is probably some multiple of x 2 − x y + y 2 x^2-xy+y^2 . Since x 2 − x y + y 2 x^2-xy+y^2 is a factor of x 3 + y 3 x^3+y^3 and I noticed that x 3 + y 3 = 3 2 x^3+y^3 = 3^2 , I started to play with ( x − y ) ( x + y ) \sqrt{(x-y)}(x+y) . It turns out it has a nice form x v − 1 xv-1 even though it wasn't I originally expected (I expected it is z 3 \sqrt[3]{z} for some z z ). The rest is just be brave to multiply out ( x v − 1 ) ( x 2 − x y + y 2 ) (xv-1)(x^2-xy+y^2) to see what's ahead, it turns out it's exactly the same form as the RHS. Hope this helps.

George G - 7 years, 5 months ago

I'm very sorry George, but I don't seem to get how all of this you wrote leads us to a ∨ b = 2 ∨ 20 a \vee b = 2 \vee 20 and c = 25 c = 25 .

This problem is very similar to "3, 2, 1, liftoff", which has been featured some months ago during the old Brilliant form. I remember the user Mark H. wrote a very beautiful general analysis, for any two values (in this case they are 5 5 and 4 4 ) inside the radical, but I can't find the link.

Guilherme Dela Corte - 7 years, 5 months ago

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Plug in the values of x , y , v x,y,v into the last expression v + x y − x 2 v+xy-x^2 .

George G - 7 years, 5 months ago

Can you please explain please that why you all used 9 at the place of 3, as it is given in question...🙄

Himanshu Jha - 11 months, 3 weeks ago
Patrick Corn
Dec 22, 2013

Square both sides to get 9 5 3 − 9 4 3 = a 2 3 + b 2 3 + c 2 3 + 2 a b 3 − 2 a c 3 − 2 b c 3 . 9 \sqrt[3]{5} - 9 \sqrt[3]{4} = \sqrt[3]{a^2} + \sqrt[3]{b^2} + \sqrt[3]{c^2} + 2 \sqrt[3]{ab} - 2 \sqrt[3]{ac} - 2 \sqrt[3]{bc}. It's going to be difficult to have all the cube roots on the right side be integer multiples of 5 3 \sqrt[3] 5 or 4 3 \sqrt[3]{4} , so we'll want two of those terms to cancel. Let's make it the second and fifth. So b 2 = 8 a c b^2 = 8ac . Then the right side becomes a 2 3 + b 4 a a b 3 + 2 a b 3 − b a a 2 3 \sqrt[3] {a^2} + \frac{b}{4a} \sqrt[3]{ab} + 2\sqrt[3]{ab} - \frac{b}{a} \sqrt[3]{a^2} , which becomes ( 1 − b a ) a 2 3 + ( 2 + b 4 a ) a b 3 . \left( 1-\frac{b}{a} \right) \sqrt[3]{a^2} + \left( 2 + \frac{b}{4a} \right) \sqrt[3]{ab}.

At this point we can probably try harder to derive the values of a a and b b , but just trying a = 2 a = 2 leads us to 1 − b / 2 = − 9 1-b/2 = -9 and b = 20 b = 20 . It's easy to check that this works. So c = b 2 / 8 a = 25 c = b^2/8a = 25 , and the sum is 47 \fbox{47} .

6 Helpful 1 Interesting 1 Brilliant 0 Confused

When you say "Let's make it the second and fifth.", how did you decide to make it those values?

minimario minimario - 7 years, 5 months ago

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Well, we could also have made it the first and last, but it doesn't matter, since the expression is symmetric in a and b.

Kevin Sun - 7 years, 5 months ago
Rohit Bhagwat
Jan 30, 2014

I dont know how to solve this analatically but a small bit of program can serve as a good tool for this one! this is how i did it in python 27. N1=1 n2=100 d=1.0503209282769135 #value for LHS of given equation eps=1.*10 (-6) def f(x): return x (1/3.) for i in range(N1,n2): for j in range(N1,n2): for k in range(N1,n2): p=f(i) q=f(j) r=f(k) if (d-eps)<abs(p+q-r) < (d+eps): print i,j,k

1 Helpful 0 Interesting 0 Brilliant 1 Confused

I don' know how to make you understand this analytically but a small bit of program can serve as a good tool for this one!

int main()

{cout<<"Stop posting programming answers in math section\n";

cout<<"It doesn't help anyone\n";

return 0;

}

Bruce Wayne - 7 years, 4 months ago

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Although this is a bit harsh, I must admit that the "return 0" thing works out pretty well here.

Harvey Dent - 7 years, 3 months ago

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