Solve the equation involving binomial theorem

[chwala]

Homework Statement

Solve the equation; $(7-6x)^3+(7+6x)^3=1736$

Relevant Equations

binomial theorem

$$(7-6x)^3+(7+6x)^3=1736$$
$$⇒(7^3(1-\frac {6}{7}x)^3+(7^3(1+\frac {6}{7}x)^3=1736$$
$$343[1-\frac {18}{7}x+\frac {216}{98}x^2-\frac{1296}{2058}x^3]+343[1+\frac {18}{7}x+\frac {216}{98}x^2+\frac{1296}{2058}x^3]=1736$$
$$343[2+\frac {432}{98}x^2]=1736$$
$$686+\frac {148,176}{98}x^2=1736$$
$$\frac {148,176}{98}x^2=1050$$
$$148,176x^2=102,900$$
$$x^2=\frac {102,900}{148,176}$$
$$x^2=0.69444$$
$$x=±0.8333$$
you can imagine the number of times i have gone through this problem, looking for an error in the expansion...only to realize that i had not brought in the factorials...lol

 

[anuttarasammyak]

Why do not you apply the formula
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$?

 

[chwala]

Why do not you apply the formula
$$a^3+b^3=(a+b)(a^2-ab+b^2)$$?

I am used to my way of expanding...been using that for years...thanks...

 

[anuttarasammyak]

In my way I have got another solution of simple fraction. I should appreciate it if you would check your answer.

 

[chwala]

In my way I have got another solution than yours. I should appreciate it if you would check your answer.

What solution did you get? if you substitute the solution to its original equation then you would confirm that it satisfies the problem...

 

[anuttarasammyak]

The equation becomes in my way
$$\frac{1736}{14}=49+3*36x^2$$

 

[chwala]

The equation becomes in my way
$$\frac{1736}{14}=49+3*36x^2$$

Which will in turn give you the same solution as the one I found.

 

[anuttarasammyak]

Your x^2 is more complex fraction than mine. I hesitate to write it down because of homework policy.

 

[chwala]

I do not seem to understand/ get you, I just checked your working and the two solutions are equivalent.

 

[chwala]

you have from your post $6$,

...$$\frac {75}{108}=x^2$$

$$0.69444=x^2$$ which is the same as what i had found ...now can you take square roots on both sides to find the value of $x?$

 

[anuttarasammyak]

75/108 is further reducible and I am afraid it does not equal to 102900/148176.

 

[chwala]

Interesting, then what is your final solution? I think i will leave it at here...and wait for other members to give their views. Cheers mate.

 

[anuttarasammyak]

I showed the equation in #7 which is solved easily. I cannot show the solution explicitly due to homework policy.

 

[chwala]

You mean $$x= \sqrt {\frac {75}{108}}$$
$$x= \sqrt {\frac {25}{36}}= ±\frac {5}{6} =±0.833333$$

 

[anuttarasammyak]

Yes. Does it coincide with your result ?

 

[chwala]

Yes. Does it coincide with your result ?

Yes, why not? ...From my post $1$,

$$\frac {102,900}{148,176}≡\frac {25}{36} $$ {divide numerator and denominator by $4116$}
$$⇒x^2=\frac {102,900}{148,176}$$ or
$$⇒x^2=\frac {25}{36}$$

 

[anuttarasammyak]

My bad in factorization,
$$102900/148176 =\frac{2*3*5^2*7^3}{2^3*3^3*7^3}=(\frac{5}{6})^2$$