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Note by Saran Balachandar 2 years, 8 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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

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Can you solve it??

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

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

@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 \( f(x) = x^3 - 7x^2 - 6x + 5 \).

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

What have you tried? Where are you stuck?

Thanks ! Got It.

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

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestHint:- \( (a+b)(b+c)(c+a) = (a+b+c)(ab+cb+ca) - abc \)

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Can you solve it??

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

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

Log in to reply

@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 \( f(x) = x^3 - 7x^2 - 6x + 5 \).

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

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What have you tried? Where are you stuck?

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Thanks ! Got It.

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

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