Question

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

• A. $0$
• B. $1$
• C. $10$
• D. $30$

Answer 1

The correct option is C $10$

We know that one of the identity of polynomials is:

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Now let us substitute $a=1.5,b=4.7$ and $c=3.8$ in the above identity as follows:

$(1.5{)}^3+(4.7{)}^3+(3.8{)}^3-3\times 1.5\times 4.7\times 3.8$

$=(1.5+4.7+3.8)\left\{\right(1.5{)}^2+(4.7{)}^2+(3.8{)}^2-(1.5\times 4.7)$

$-(4.7\times 3.8)-(1.5\times 3.8)\}$ .....(1)

Now consider the given expression as shown below:

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

$=\frac{(1.5+4.7+3.8)\left\{\right(1.5{)}^2+(4.7{)}^2+(3.8{)}^2-(1.5\times 4.7)-(4.7\times 3.8)-(1.5\times 3.8)\}}{(1.5{)}^2+(4.7{)}^2+(3.8{)}^2-(1.5\times 4.7)-(4.7\times 3.8)-(1.5\times 3.8)}$ ....From eq(1)

$=1.5+4.7+3.8=10$