Calculating the Value of $99^3$
To find the value of $99^3$, we can use the algebraic identity for the cube of a binomial difference: $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$.
We rewrite $99$ as $(100 - 1)$. In this case, $a = 100$ and $b = 1$.
Applying the Formula
- Term 1 ($a^3$): Calculate $100^3$. $100^3 = 100 \times 100 \times 100 = 1,000,000$
- Term 2 ($-3a^2b$): Calculate $-3 \times (100)^2 \times 1$. $-3 \times 100^2 \times 1 = -3 \times 10,000 \times 1 = -30,000$
- Term 3 ($+3ab^2$): Calculate $+3 \times 100 \times (1)^2$. $+3 \times 100 \times 1^2 = +3 \times 100 \times 1 = +300$
- Term 4 ($-b^3$): Calculate $-1^3$. $-1^3 = -1$
Final Calculation
Sum the results from the previous step:
$99^3 = 1,000,000 - 30,000 + 300 - 1$
$99^3 = 970,000 + 300 - 1$
$99^3 = 970,300 - 1$
$99^3 = 970,299$
The value of $99^3$ is $970,299$.
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Important Questions from Cube and Cube Root
If 5 \(\sqrt{3}\) + \(\sqrt{75}\) = 17.32, then the value of 14 \(\sqrt{3}\) + \(\sqrt{108}\) is:
Find the value of (25)3 + (-29)3 + (4)3
The value of \(\frac{\sqrt[3]{-2744} \times \sqrt[3]{-216}}{\sqrt[3]{\frac{64}{729}}}\) is:
A number is cube of 53. When 7 times of 57 is subtracted from the number, then the resultant number which is formed will be divisible by:
The sum of a positive number and its cube is 1740. What is the value of the number?
