MHB Solving an Equation with Fractions

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The discussion focuses on solving the equation (x^3 + 3x)/341 = (3x^2 + 1)/91 using properties of proportions. Various methods are presented, including the use of the relationship between cubes and the application of the componendo dividendo technique. The problem is simplified by recognizing that 341 and 91 can be expressed as sums and differences of cubes, leading to the equation 216(x-1)^3 = 125(x+1)^3. Ultimately, the solution yields x = 11 through different approaches, confirming the validity of the methods used. The thread highlights the versatility of algebraic techniques in solving equations involving fractions.
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using properties of proportion solve

( x^3 + 3x )/341 = (3x^2 + 1)/91
 
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kuheli said:
using properties of proportion solve

( x^3 + 3x )/341 = (3x^2 + 1)/91

If $\displaystyle f(x) = (1+x)^{3}$ then is...

$\displaystyle \text{Even} \{f(x)\} = \frac{f(x) + f(-x)}{2}$

$\displaystyle \text{Odd} \{f(x)\}= \frac{f(x)-f(-x)}{2}\ (1)$

... so that with little effort we obtain... $\displaystyle 125\ f(x) = - 216\ f(-x) \implies 125\ (1+x)^{3}= - 216\ (1-x)^{3} \implies \frac{(1 + x)^{3}}{(1-x)^{3}} = - \frac{216}{125} \implies \frac{1+x}{1-x} = (- \frac {216}{125})^{\frac{1}{3}}\ = - \frac{6}{5}\ (2)$

Now all what we have to do is to solve a first order equation... Kind regards $\chi$ $\sigma$
 
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Here is another method to approach the problem:

The existence of the terms $x^3+3x$ and $3x+1$ (as given in the problem) suggests that we might want to consider to relate the problem with $(x\pm1)^3$ and also, 341 and 91 be the sum or difference of cubes.

Observe that $341=216+125$ and $91=216-125$, hence we have

$$\frac{x^3+3x}{341}=\frac{3x^2+1}{91}$$

$$\frac{x^3+3x}{216+125}=\frac{3x^2+1}{216-125}$$

$$216(x^3+3x)-125(x^3+3x)=216(3x^2+1)+125(3x^2+1)$$

$$216(x^3+3x)-216(3x^2+1)=125(3x^2+1)+125(x^3+3x)$$

$$216(x^3-3x^2+3x-1)=125(x^3+3x^2+3x+1)$$

$$216(x-1)^3=125(x+1)^3$$

$$\frac{(x-1)^3}{(x+1)^3}=\frac{125}{216}=\frac{5^3}{6^3}$$

$$\frac{x-1}{x+1}=\frac{5}{6}$$

$$\therefore x=11$$
 
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I feel a little unnecessary in providing an answer when there are already 2 good answers but because my method is different I would provide it

we have

$\frac{x^3 + 3x}{341} = \frac{3x^2 + 1}{91}$

which is same as

$\frac{x^3 + 3x}{3x^2 + 1} = \frac{341}{91}$

using componendo dividendo we have

$\frac{x^3 + 3x+3x^2 + 1}{x^3 + 3x -3x^2 - 1} = \frac{341+91}{341- 91}$

rearranging the terms on LHS and simplifying RHS we get

$\frac{x^3 + 3x^2+3x + 1}{x^3 - 3x^2+ 3x - 1} = \frac{432}{250}$

or
$\frac{(x+1)^3}{(x-1)^3} = \frac{216}{125}$

or $(\frac{x+1}{x-1})^3 = (\frac{6}{5})^3$

or $\frac{x+1}{x-1} = \frac{6}{5}$

I apply componendo dividendo again

$\frac{x+1+x - 1}{(x+1)- (x-1)} = \frac{6+5 }{6- 5}$
or
$\frac{2x}{2} = \frac{11 }{1}$

or x = 11
for componendo dividendo method if some is unfamiliar

we have if

$\frac{a}{b} = \frac{c }{d}$

then

$\frac{a+b}{a-b} = \frac{c +d }{c- d}$
 
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