Question

(a-b)^2 -c^2/a^2-(b+c)^2+(b-c)^2-a^2/b^2-(c+a)^2+(c-a)^2-b^2/c^2-(a+b)^2​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{\displaystyle\frac{\left(a-b\right)^2\:-c^2}{a^2-\left(b+c\right)^2}+\frac{\left(b-c\right)^2-a^2}{b^2-\left(c+a\right)^2}+\frac{\left(c-a\right)^2-b^2}{c^2-\left(a+b\right)^2}}$

★═════════════════★  

♣ ᴀɴꜱᴡᴇʀ :

$\sf{\displaystyle\frac{\left(a-b\right)^2\:-c^2}{a^2-\left(b+c\right)^2}+\frac{\left(b-c\right)^2-a^2}{b^2-\left(c+a\right)^2}+\frac{\left(c-a\right)^2-b^2}{c^2-\left(a+b\right)^2}=1}$

★═════════════════★

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sf{\displaystyle\frac{\left(a-b\right)^2\:-c^2}{a^2-\left(b+c\right)^2}+\frac{\left(b-c\right)^2-a^2}{b^2-\left(c+a\right)^2}+\frac{\left(c-a\right)^2-b^2}{c^2-\left(a+b\right)^2}}$

$\bigstar\:\:\sf{\text { Cancel } \dfrac{(a-b)^{2}-c^{2}}{a^{2}-(b+c)^{2}}: \quad \dfrac{a+c-b}{a+b+c}}$

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

$\bigstar\:\:\sf{\displaystyle\text { Cancel } \frac{(b-c)^{2}-a^{2}}{b^{2}-(c+a)^{2}}: \quad \frac{a+b-c}{a+b+c}}$

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

$\bigstar\:\:\sf{\displaystyle\text { Cancel }\frac{(c-a)^{2}-b^{2}}{c^{2}-(a+b)^{2}}: \quad \frac{b+c-a}{a+b+c}}$

$\sf{\displaystyle=\frac{a+c-b}{a+b+c}+\frac{a+b-c}{a+b+c}+\frac{b+c-a}{a+b+c}}$

$\bf{\mathrm{Apply\:rule}\:\dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}}$

$\sf{\displaystyle=\frac{a-b+c+b-c+a+c-a+b}{a+b+c}}$

$\sf{=\dfrac{a+b+c}{a+b+c}}$

$\Large\boxed{\sf{=1}}$