Question from AMTI 2015

#NumberTheory

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Comments

Hint:- $(a+b)(b+c)(c+a) = (a+b+c)(ab+cb+ca) - abc$

Siddhartha Srivastava - 5 years, 8 months ago

Can you solve it??

Saran Balachandar - 5 years, 8 months ago

Using the hint given by Siddhartha Srivatsava, you can substitute the values of

$(a+b+c)(ab+cb+ca) - abc$ by Vieta's formula

A Former Brilliant Member - 5 years, 8 months ago

@Siddhartha Srivastava Nice hint. Here is another approach which doesn't require knowing the algebraic identity.

$(a+b)(a+c)(b+c) = ( s - a)(s-b) ( s-c) = f(s)$ where $s = a+b+c$ and $f(x) = x^3 - 7x^2 - 6x + 5$ .

Calvin Lin Staff - 5 years, 8 months ago

What have you tried? Where are you stuck?

Calvin Lin Staff - 5 years, 8 months ago

Thanks ! Got It.

Saran Balachandar - 5 years, 8 months ago

$If\quad a\quad cubic\quad polynomial\quad P(x)=ax^{ 3 }+bx^{ 2 }+cx+d\quad satisifies\quad the\quad condtition\quad P(X)=0,\quad \\ then\quad the\quad roots\quad x_{ 1 },x_{ 2 },x_{ 3 }\quad of\quad the\quad equation\quad P(x)=0\quad satisfy\quad x_{ 1 }+x_{ 2 }+x_{ 3 }=-\frac { b }{ a } ,\quad \\ x_{ 1 }x_{ 2 }+x_{ 1 }x_{ 3 }+x_{ 2 }x_{ 3 }=\frac { c }{ a } ,\quad x_{ 1 }x_{ 2 }x_{ 3 }=-\frac { d }{ a } .\\ (a+b)(b+c)(c+a)\quad can\quad be\quad written\quad as\quad (a+b+c)(ab+bc+ca)-abc\\ \Rightarrow \quad (7)(-6)-(-5)\quad \Rightarrow \quad -37$

Saran Balachandar - 5 years, 8 months ago

{7x (-6)}-(-5)=-42+5= -37

Pranjal Prashant - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$