Question

a-b)(a + b) + (1 - 0) (b+c) + (c-a) (c+a)=0​

Answer 1

Answer:

ab+bc+ca=0

bc=−ca−ab

bc=−a(b+c)---eq1

like wise,

ab=−c(a+b)---eq2

& ca=−b(c+a)---eq3

=a2−bc1+b2−ca1+c2−ab1

taking the value from eq1, 2&3

=a2+a(b+c)1+b2+b(a+c)1+c2+c(a+b)1

=a(a+b+c)1+b(a+b+c)1+c(a+b+c)1

=abc(a+b+c)ab+bc+ca

As ab+bc+ca=0

So, =0

Answer 2

Step-by-step explanation:

$(a-b)(a+b)+(1-0)(b+c)+(c-a)(c+a)=0$

$a^{2}-b^{2}+\left(1-0\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)=0$

$a^{2}-b^{2}+1\left(b+c\right)+\left(c-a\right)\left(c+a\right)=0$

$a^{2}-b^{2}+b+c+\left(c-a\right)\left(c+a\right)=0$

$a^{2}-b^{2}+b+c+c^{2}-a^{2}=0$

$-b^{2}+b+c+c^{2}=0$

$-b^{2}+b+c^{2}+c=0$

$b=\frac{-1±\sqrt{1^{2}-4\left(-1\right)c\left(c+1\right)}}{2\left(-1\right)}$

$b=\frac{-1±\sqrt{1-4\left(-1\right)c\left(c+1\right)}}{2\left(-1\right)}$

$b=\frac{-1±\sqrt{1+4c\left(c+1\right)}}{2\left(-1\right)}$

$b=\frac{-1±\sqrt{\left(2c+1\right)^{2}}}{2\left(-1\right)}$

$b=\frac{-1±|2c+1|}{2\left(-1\right)}$

$b=\frac{-1±|2c+1|}{-2}$

$b=\frac{|2c+1|-1}{-2}$

$b=\frac{-|2c+1|+1}{2}$

$b=\frac{-|2c+1|-1}{-2}$

$b=\frac{|2c+1|+1}{2}$

$b=\frac{-|2c+1|+1}{2} b=\frac{|2c+1|+1}{2}$

hope it helps you ☺️☺️☺️