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# Question from AMTI 2015

Note by Saran Balachandar
1 year, 11 months ago

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Hint:- $$(a+b)(b+c)(c+a) = (a+b+c)(ab+cb+ca) - abc$$ · 1 year, 11 months ago

Can you solve it?? · 1 year, 11 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 · 1 year, 11 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 $$f(x) = x^3 - 7x^2 - 6x + 5$$. Staff · 1 year, 11 months ago

$$If\quad a\quad cubic\quad polynomial\quad P(x)=ax^{ 3 }+bx^{ 2 }+cx+d\quad satisifies\quad the\quad condtition\quad P(X)=0,\quad \\ then\quad the\quad roots\quad x_{ 1 },x_{ 2 },x_{ 3 }\quad of\quad the\quad equation\quad P(x)=0\quad satisfy\quad x_{ 1 }+x_{ 2 }+x_{ 3 }=-\frac { b }{ a } ,\quad \\ x_{ 1 }x_{ 2 }+x_{ 1 }x_{ 3 }+x_{ 2 }x_{ 3 }=\frac { c }{ a } ,\quad x_{ 1 }x_{ 2 }x_{ 3 }=-\frac { d }{ a } .\\ (a+b)(b+c)(c+a)\quad can\quad be\quad written\quad as\quad (a+b+c)(ab+bc+ca)-abc\\ \Rightarrow \quad (7)(-6)-(-5)\quad \Rightarrow \quad -37$$ · 1 year, 11 months ago

What have you tried? Where are you stuck? Staff · 1 year, 11 months ago

Thanks ! Got It. · 1 year, 11 months ago