Question

The value of $\frac{(1.5{)}^3+(4.7{)}^3+(3.8{)}^3-3\times 1.5\times 4.7\times 3.8}{(1.5{)}^2+(4.7{)}^2+(3.8{)}^2-1.5\times 4.7-4.7\times 3.8-1.5\times 3.8}$ is

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

Answer: (C)

$10$

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

$=$ $\frac{\frac{1}{2}(a+b+c)\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]}{\frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]}=(a+b+c)$
For $a=1.5,b=4.7,c=3.8$ value is $1.5+4.7+3.8=10$