Find $x$, if $2(a-b) + x(b-c)^2 + (c-a) ^3 = 2(a-d) + (b-d) ^2 +(c-d) ^3$, where $a, b, c, d$ are in AP series

Question

If $$2(a-b) + x(b-c)^2 + (c-a) ^3 = 2(a-d) + (b-d) ^2 +(c-d) ^3$$ where $a, b, c, d$ are four distinct real numbers and they are in Arithmetic Progression, then the possible value of $x$ can be:

(A) $-9$

(B) $-7$

(C) $14$

(D) $17$

Edit: Using help from the answer given by @NL628 I could figure this out.

Let $a-b=k$ ,

Then, $\mapsto 2k + x \cdot k^2 + (-2k)^3 = 2 \cdot (3k) + (2k)^2 + k^3$

Solving the quadratic equation for $k$ :

$$k = \frac{x-4 \pm \sqrt{x^2-8x-128}}{18}$$

What can I do now to find the value of $x$ ? What am I missing here?

Answer 1

Let $a - b = k.$ Thus, $b - c = k$ and $c - d = k$ because a, b, c, d are in arithmetic progression. Similarly, you can find out values for $a - d$ and $b - d.$

We see that the following equation simplifies to $$2k + x \cdot k^2 + (-2k)^3 = 2 \cdot (3k) + (2k)^2 + k^3$$.

With the single variable equation, I'll leave it to you to figure out the rest :)

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