$\large \left(\dfrac{a^2+b^2}{c^2}\right)\left[\left(\dfrac{a^3+b^3+c^3}{abc}\right)^6 - 29\right]$

Consider a polynomial such that its roots are in arithmetic progression and the squares of these roots are also in an arithmetic progression in the same order.

If the roots are $a,b,c$ such that $a,b,c \neq 0$ , find the value of the said expression above.

The answer is 1400.

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Awesome question and excellent explanation.

Gagan Jain
- 6 years, 1 month ago

Let me post my
*
bashy
*
way in the comments just to show a different
*
non-elegant
*
approach.

My method is identical to this one till the part where he proves $b^2=ac$ implying that $a,b,c$ forms a geometric progression. As such, we use the general form of GP to reduce the expression (noting the homogeneity of the expression and that $a,b,c\neq 0$ ). Take $r$ as the common ratio of the GP sequence $\{a,b,c\}$ and the required expression as $E$ . We get,

$E=\frac{1+r^2}{r^4}\left(\frac{(1+r^3+r^6)^6}{r^{24}}-29\right)$

Now, from the fact that both the sequences $(a,b,c)$ and $(a^2,b^2,c^2)$ are ordered AP sequences, we get two equations:

$\begin{cases}1+r^2=2r\\ 2r^2=1+r^4\end{cases}\implies r+r^3=2r^2=1+r^4\implies r^4-r^3-r+1=0$

This polynomial equation has only one real solution, which is
$r=1$
. Now, simply substitute this value of
$r$
into the
*
reduced expression
*
for
$E$
and you get,

$E=\frac{1+1}{1}\left(\frac{3^6}{1}-29\right)=1400$

**
Note:
**
By
*
ordered
*
AP sequence, I'm referring to the fact that the order of the AP sequence is as written there and cannot be permuted randomly. It can only be permuted in at most another way: reversely. (last and first elements switch)

Prasun Biswas
- 6 years, 1 month ago

How about $(a, b, c)=(7, 1, -5)$ ?

Joel Tan
- 6 years, 1 month ago

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$(a^2,b^2,c^2)=(49,1,25)\implies a^2,c^2,b^2\textrm{ are in GP}\not\Rightarrow a^2,b^2,c^2\textrm{ are in GP}$

$\not\Rightarrow$ denotes "Not implies"

I agree that the problem needs to be rephrased a bit to mention the ordering.

Prasun Biswas
- 6 years, 1 month ago

$a^2,b^2, c^2$ is not in A.P.

Niranjan Khanderia
- 6 years, 1 month ago

OMG Dude! This didn't strike me! Good one indeed. Perfect. "One of the finest minds in the country."

Mehul Arora
- 6 years, 1 month ago

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Thanks hawkeye .. xD

Nihar Mahajan
- 6 years, 1 month ago

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This was quite an easy problem.

Ronak Agarwal
- 6 years, 1 month ago

Cool method.but after knowing that a,b,c are in AP as well as in GP we can write a=b=c. So substituting in eqn we get the ans 1400

prakhar mishra
- 5 years, 11 months ago

i took a, c, b in ap

Dev Sharma
- 5 years, 6 months ago

We know: b-a = c - b

and

b²-a² = c²-b² ->

-> (b-a)(b+a) = (c-b)(c+b) ->

-> a+b = c+b ->

a = b = c

The expression is equal to 2. (729-9) = 1400

2 Helpful
0 Interesting
0 Brilliant
0 Confused

Dividing $b-a$ in the equation $(b-a)(b+a)=(c-b)(c+b) \Longrightarrow b \neq a$ , resulting in a contradiction.

Shak R
- 6 years, 1 month ago

1 Helpful
0 Interesting
0 Brilliant
0 Confused

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According to the given condition ,

$a+c=2b\dots (1) \\ a^2+c^2=2b^2\dots (2) \\ \Rightarrow (a+c)^2-2ac=2b^2 \\ \Rightarrow (2b)^2-2ac = 2b^2 \dots \text{from (1)}\\ \Rightarrow 2ac = 2b^2 \\ \Rightarrow b^2 = ac \dots (3)$

Note:We get an astonishing result that $a,b,c$ also form a geometric progression!Substituting $(3)$ in $(2)$ ,

$a^2+c^2=2ac \\ \Rightarrow (a-c)^2 = 0 \\ \Rightarrow a=c \dots (4)$

Substituting $(4)$ in $(3)$ ,

$b^2 = a^2 \\ \Rightarrow a=b \\ \Rightarrow a=b=c \quad \text{(by transitivity)}$

This makes the given expression:

$\left(\dfrac{2a^2}{a^2}\right)\left(\left(\dfrac{3a^3}{a^3}\right)^6-29\right) \\ = (2)(3^6-29)\\ =2(729-29)\\ =2(700) \\ =\huge\boxed{\color{#3D99F6}{1400}}$