An algebra problem by madhav srirangan

Algebra Level 5

729 x 6 − 1458 x 5 − 1215 x 4 − 540 x 3 − 135 x 2 − 18 x − 1 = 0 729x^{6}-1458x^{5}-1215x^{4}-540x^{3}-135x^{2}-18x-1=0

The above equation has a positive real root in the form

1 3 ( a 1 / 6 + b 1 / 6 + c 1 / 6 + d 1 / 6 + e 1 / 6 + f 1 / 6 ) . \frac13(a^{1/6}+b^{1/6}+c^{1/6}+d^{1/6}+e^{1/6}+f^{1/6}).

with a , b , c , d , e , f a, b, c, d, e, f are all integers. Find a + b + c + d + e + f + 3 a+b+c+d+e+f+3 .


The answer is 66.

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Madhav Srirangan by madhav srirangan , 20, India

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3 solutions

Send all the negative terms to the RHS:

729 x 6 = 1458 x 5 + 1215 x 4 + 540 x 3 + 135 x 2 + 18 x + 1 729x^6=1458x^5+1215x^4+540x^3+135x^2+18x+1

Add 729 x 6 729x^6 to both sides:

1458 x 6 = 729 x 6 + 1458 x 5 + 1215 x 4 + 540 x 3 + 135 x 2 + 18 x + 1 1458x^6=729x^6+1458x^5+1215x^4+540x^3+135x^2+18x+1

Factor the RHS:

1458 x 6 = ( 3 x + 1 ) 6 1458x^6=(3x+1)^6

Since we only want the positive real solution:

1458 x 6 6 = ( 3 x + 1 ) 6 6 3 2 6 x = 3 x + 1 3 ( 2 6 − 1 ) x = 1 3 x = 1 2 6 − 1 \sqrt[6]{1458x^6}=\sqrt[6]{(3x+1)^6} \\ 3\sqrt[6]{2}x=3x+1 \\ 3(\sqrt[6]{2}-1)x=1 \\ 3x=\dfrac{1}{\sqrt[6]{2}-1}

After rationalizing the denominator we get:

3 x = 32 6 + 16 6 + 8 6 + 4 6 + 2 6 + 1 6 2 − 1 3x=\dfrac{\sqrt[6]{32}+\sqrt[6]{16}+\sqrt[6]{8}+\sqrt[6]{4}+\sqrt[6]{2}+\sqrt[6]{1}}{2-1}

Finally, solve for x x :

x = 32 6 + 16 6 + 8 6 + 4 6 + 2 6 + 1 6 3 x=\dfrac{\sqrt[6]{32}+\sqrt[6]{16}+\sqrt[6]{8}+\sqrt[6]{4}+\sqrt[6]{2}+\sqrt[6]{1}}{3}

Comparing we get a = 32 , b = 16 , c = 8 , d = 4 , e = 2 , f = 1 a=32,b=16,c=8,d=4,e=2,f=1 , so a + b + c + d + e + f + 3 = 66 a+b+c+d+e+f+3=\boxed{66}

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Chew-Seong Cheong
Aug 24, 2015

729 x 6 − 1458 x 5 − 1215 x 4 − 540 x 3 − 135 x 2 − 18 x − 1 = 0 ( 3 x ) 6 − 6 ( 3 x ) 5 − 15 ( 3 x ) 4 − 20 ( 3 x ) 3 − 15 ( 3 x ) 2 − 6 ( 3 x ) − 1 = 0 y 6 − 6 y 5 − 15 y 4 − 20 y 3 − 15 y 2 − 6 y − 1 = 0 2 y 6 − ( y + 1 ) 6 = 0 2 y 6 = ( y + 1 ) 6 2 1 6 y = y + 1 \begin{aligned} 729x^6-1458x^5-1215x^4-540x^3-135x^2-18x-1 & = 0 \\ (3x)^6-6(3x)^5-15(3x)^4-20(3x)^3-15(3x)^2-6(3x)-1 & = 0 \\ y^6-6y^5-15y^4-20y^3-15y^2-6y-1 & = 0 \\ 2y^6 - (y+1)^6 & = 0 \\ 2y^6 & = (y+1)^6 \\ 2^{\frac{1}{6}}y & = y + 1 \end{aligned}

⇒ y = 1 2 1 6 − 1 = 1 + 2 1 6 + 2 2 6 + 2 3 6 + 2 4 6 + 2 5 6 3 x = 1 + 2 1 6 + 4 1 6 + 8 1 6 + 1 6 1 6 + 3 2 1 6 ⇒ x = 1 + 2 1 6 + 4 1 6 + 8 1 6 + 1 6 1 6 + 3 2 1 6 3 \begin{aligned} \Rightarrow y & = \frac{1}{2^{\frac{1}{6}}-1} \\ & = 1 + 2^{\frac{1}{6}} + 2^{\frac{2}{6}} + 2^{\frac{3}{6}} + 2^{\frac{4}{6}} + 2^{\frac{5}{6}} \\ 3x & = 1 + 2^{\frac{1}{6}} + 4^{\frac{1}{6}} + 8^{\frac{1}{6}} + 16^{\frac{1}{6}} + 32^{\frac{1}{6}} \\ \Rightarrow x & = \frac{1 + 2^{\frac{1}{6}} + 4^{\frac{1}{6}} + 8^{\frac{1}{6}} + 16^{\frac{1}{6}} + 32^{\frac{1}{6}}}{3} \end{aligned}

⇒ a + b + c + d + e + f = 32 + 16 + 8 + 4 + 2 + 1 + 3 = 66 \Rightarrow a + b + c + d + e + f = 32+16+8+4+2+1+3 = \boxed{66}

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Moderator note:

It's great that you've identified the elements on the 6th row of the Pascal Triangle. Fantastic observation!

And of course, for completeness, can you show that there is only one positive root?

@Chew-Seong Cheong , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Brilliant Mathematics Staff - 5 years, 9 months ago

2^(1/ 6) y = y + 1 has been a step as the master mind for an answer wanted.

Lu Chee Ket - 5 years, 8 months ago
Lu Chee Ket
Sep 27, 2015

Only reals are 2.7219317162754064264+ and -0.15705031503335800006+, with Wolfram Alpha. This question required good observational power to solve particularly. By hard and general way, a, b, c, d, e, and f can be searched successfully by all presumed distinct initially. 2.7219317162754064264 taken as constant found with 1, 2, 4, 8, 16 and 32. Plus 3 equals to 66.

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