Solve (b+c)2/3bc+(a+c)2/3ac+(b+a)2/3ba=1

$$\frac { ( b + c ) 2 } { 3 b c } + \frac { ( a + c ) 2 } { 3 a c } + \frac { ( b + a ) 2 } { 3 b a } = 1$$

Solve for a

$$a=-\frac{4bc}{4c+4b-3bc}$$

$$c\neq 0\text{ and }b\neq 0\text{ and }\left(c=\frac{4}{3}\text{ or }b\neq -\frac{4c}{4-3c}\right)$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3abc, the least common multiple of 3bc,3ac,3ba.

$$a\left(b+c\right)\times 2+b\left(a+c\right)\times 2+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply a by b+c.

$$\left(ab+ac\right)\times 2+b\left(a+c\right)\times 2+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply ab+ac by 2.

$$2ab+2ac+b\left(a+c\right)\times 2+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply b by a+c.

$$2ab+2ac+\left(ba+bc\right)\times 2+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply ba+bc by 2.

$$2ab+2ac+2ba+2bc+c\left(b+a\right)\times 2=3abc$$

Combine 2ab and 2ba to get 4ab.

$$4ab+2ac+2bc+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply c by b+a.

$$4ab+2ac+2bc+\left(cb+ca\right)\times 2=3abc$$

Use the distributive property to multiply cb+ca by 2.

$$4ab+2ac+2bc+2cb+2ca=3abc$$

Combine 2bc and 2cb to get 4bc.

$$4ab+2ac+4bc+2ca=3abc$$

Combine 2ac and 2ca to get 4ac.

$$4ab+4ac+4bc=3abc$$

Subtract 3abc from both sides.

$$4ab+4ac+4bc-3abc=0$$

Subtract 4bc from both sides. Anything subtracted from zero gives its negation.

$$4ab+4ac-3abc=-4bc$$

Combine all terms containing a.

$$\left(4b+4c-3bc\right)a=-4bc$$

The equation is in standard form.

$$\left(4c+4b-3bc\right)a=-4bc$$

Divide both sides by 4c+4b-3cb.

$$\frac{\left(4c+4b-3bc\right)a}{4c+4b-3bc}=-\frac{4bc}{4c+4b-3bc}$$

Dividing by 4c+4b-3cb undoes the multiplication by 4c+4b-3cb.

$$a=-\frac{4bc}{4c+4b-3bc}$$

Variable a cannot be equal to 0.

$$a=-\frac{4bc}{4c+4b-3bc}\text{, }a\neq 0$$

Solve for b

$$b=-\frac{4ac}{4c+4a-3ac}$$

$$c\neq 0\text{ and }a\neq 0\text{ and }\left(a=\frac{4}{3}\text{ or }c\neq -\frac{4a}{4-3a}\right)$$

Steps for Solving Linear Equation

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3abc, the least common multiple of 3bc,3ac,3ba.

$$a\left(b+c\right)\times 2+b\left(a+c\right)\times 2+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply a by b+c.

$$\left(ab+ac\right)\times 2+b\left(a+c\right)\times 2+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply ab+ac by 2.

$$2ab+2ac+b\left(a+c\right)\times 2+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply b by a+c.

$$2ab+2ac+\left(ba+bc\right)\times 2+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply ba+bc by 2.

$$2ab+2ac+2ba+2bc+c\left(b+a\right)\times 2=3abc$$

Combine 2ab and 2ba to get 4ab.

$$4ab+2ac+2bc+c\left(b+a\right)\times 2=3abc$$

Use the distributive property to multiply c by b+a.

$$4ab+2ac+2bc+\left(cb+ca\right)\times 2=3abc$$

Use the distributive property to multiply cb+ca by 2.

$$4ab+2ac+2bc+2cb+2ca=3abc$$

Combine 2bc and 2cb to get 4bc.

$$4ab+2ac+4bc+2ca=3abc$$

Combine 2ac and 2ca to get 4ac.

$$4ab+4ac+4bc=3abc$$

Subtract 3abc from both sides.

$$4ab+4ac+4bc-3abc=0$$

Subtract 4ac from both sides. Anything subtracted from zero gives its negation.

$$4ab+4bc-3abc=-4ac$$

Combine all terms containing b.

$$\left(4a+4c-3ac\right)b=-4ac$$

The equation is in standard form.

$$\left(4c+4a-3ac\right)b=-4ac$$

Divide both sides by 4c+4a-3ac.

$$\frac{\left(4c+4a-3ac\right)b}{4c+4a-3ac}=-\frac{4ac}{4c+4a-3ac}$$

Dividing by 4c+4a-3ac undoes the multiplication by 4c+4a-3ac.

$$b=-\frac{4ac}{4c+4a-3ac}$$

Variable b cannot be equal to 0.

$$b=-\frac{4ac}{4c+4a-3ac}\text{, }b\neq 0$$