Question

If $y=\frac{(0.42{)}^3+(0.25{)}^3+(0.33{)}^3-3\times 0.42\times 0.25\times 0.33}{(0.42{)}^2+(0.25{)}^2+(0.33{)}^2-0.42\times 0.25-0.25\times -0.33\times 0.42}$, then the value of $y$ is :

• A. $1$
• B. $-1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $1$

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

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

$=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{a^2+b^2+c^2-ab-bc-ca}$

$=a+b+c$

$a+b+c=0.42+0.25+0.33=1$