Question

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solve x-a(b+c)/bc+x-b(c+a)/ca+x-c(a+b)/ab = 3

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Answer 1

Answer:

x = ab + bc + ac

Step-by-step explanation:

Given :

To find x, if :

$\frac{x-a(b+c)}{bc}+\frac{x-b(c+a)}{ca}+\frac{x-c(a+b)}{ab}=3$

Solution :

$\frac{x-a(b+c)}{bc}+\frac{x-b(c+a)}{ca}+\frac{x-c(a+b)}{ab}=3$

By multiplying & dividing, a in 1st division , b in 2nd division , c in 3rd division

⇒ $\frac{x-a(b+c)}{bc} \times \frac{a}{a}+\frac{x-b(c+a)}{ca} \times \frac{b}{b}+\frac{x-c(a+b)}{ab}\times \frac{c}{c}=3$

⇒ $\frac{ax-a^2(b+c)}{abc} +\frac{bx-b^2(c+a)}{abc} +\frac{cx-c^2(a+b)}{abc}=3$

⇒ $\frac{ax-a^2(b+c)+bx-b^2(c+a)+cx-c^2(a+b)}{abc}=3$

⇒ $ax-a^2(b+c)+bx-b^2(c+a)+cx-c^2(a+b)=3abc$

⇒ $(a+b+c)x-a^2(b+c)-b^2(c+a)-c^2(a+b)=3abc$

⇒ $(a+b+c)x=a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc$

⇒ $(a+b+c)x=a^2b+a^2c+b^2c+ab^2+c^2a+c^2b+3abc$

⇒ $(a+b+c)x=a^2b+a^2c+b^2c+ab^2+c^2a+c^2b+abc+abc+abc$

⇒ $(a+b+c)x=[a^2b+ab^2+abc]+[abc+b^2c+bc^2]+[a^2c+abc+ac^2]$

⇒ $(a+b+c)x=[ab(a+b+c)]+[bc(a+b+c)]+[ac(a+b+c)]$

⇒ $(a+b+c)x=(a+b+c)[ab+bc+ac]$

⇒ $x = \frac{(a+b+c)[(ab+bc+ac)]}{a+b+c}=ab+bc+ac$

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