left(frac{10.3 times 10.3 times 10.3 + 1}{10. | vedclass

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

Question

(C) The given expression is in the form of $\frac{a^3 + b^3}{a^2 - ab + b^2}$,where $a = 10.3$ and $b = 1$.Using the algebraic identity $a^3 + b^3 = (

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) The given expression is in the form of $\frac{a^3 + b^3}{a^2 - ab + b^2}$,where $a = 10.3$ and $b = 1$.Using the algebraic identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$,we can rewrite the expression as:$\frac{(a + b)(a^2 - ab + b^2)}{(a^2 - ab + b^2)} = a + b$.Substituting the values $a = 10.3$ and $b = 1$:$10.3 + 1 = 11.3$.

$\left(\frac{10.3 \times 10.3 \times 10.3 + 1}{10.3 \times 10.3 - 10.3 + 1}\right)$ is equal to (in $.3$)

Similar Questions

What will come in the place of the question mark $(?)$ in the following equation?
$(23.1)^{2} + (48.6)^{2} - (39.8)^{2} = ? + 1147.69$

If $2994 \div 14.5 = 172$,then $29.94 \div 1.45 = ?$

The value of $\frac{(0.96)^{3}-(0.1)^{3}}{(0.96)^{2}+0.096+(0.1)^{2}}$ is

$\frac{1.49 \times 14.9 - 0.51 \times 5.1}{14.9 - 5.1}$

$\frac{1}{3} + \frac{1}{2} + \frac{1}{x} = 4$. Then $x = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module