Question

find the value of this....




plz plz plz plz plz plz plz plz plz plz plz plz plz​

Answer 1

Explanation:

$\frac{a^3+b^3+c^3-3abc}{ab+bc+ca-a^2-b^2-c^2}$

$=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{-(a^2+b^2+c^2-ab-bc-ca)}$ [applying formula directly]

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

$= \frac{-5-6+10}{-1}$ [substituting the values of a, b and c]

$=\frac{-1}{-1}$

$=1$

Answer: 1

I hope this helps. :D

Answer 2

Answer:

-125-216+1000-3(-5) (-6) (10) /30-60-50-25-36-100

-341+1000-3(300) /-151

-341+100/-151

-241/-151

241/151