frac{(0.6^{3}-0.1^{3}-0.4^{3}-3 times 0.6 tim | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $a = 0.6, b = -0.1, c = -0.4$.The given expression is of the form $\frac{a^{3} + b^{3} + c^{3} - 3abc}{a^{2} + b^{2} + c^{2} - ab - bc - ca}$.

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(B) Let $a = 0.6, b = -0.1, c = -0.4$.The given expression is of the form $\frac{a^{3} + b^{3} + c^{3} - 3abc}{a^{2} + b^{2} + c^{2} - ab - bc - ca}$.We know the algebraic identity: $a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.Substituting this into the expression,we get:$\frac{(a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)}{(a^{2} + b^{2} + c^{2} - ab - bc - ca)} = a + b + c$.Substituting the values of $a, b,$ and $c$:$0.6 + (-0.1) + (-0.4) = 0.6 - 0.1 - 0.4 = 0.1$.

$\frac{(0.6^{3}-0.1^{3}-0.4^{3}-3 \times 0.6 \times 0.1 \times 0.4)}{(0.6^{2}+0.1^{2}+0.4^{2}+0.6 \times 0.1+0.6 \times 0.4-0.1 \times 0.4)}$

Similar Questions

If $2 \frac{1}{2}$ is added to a number and the sum is multiplied by $4 \frac{1}{2}$,and $3$ is added to the product,and then dividing the sum by $1 \frac{1}{5}$,the quotient becomes $25$. What is the number?

The sum of three consecutive odd numbers is $20$ more than the first of these numbers. What is the middle number?

$\frac{753 \times 753 + 247 \times 247 - 753 \times 247}{753 \times 753 \times 753 + 247 \times 247 \times 247} = ?$

If the sum of a number and its square is $182$,what is the number?

What least number must be subtracted from $1294$ so that the remainder when divided by $9, 11, 13$ will leave in each case the same remainder $6$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module