Question
Evaluate the following: $104^3 + 96^3$
✓
Answer
Given,
$104^3 + 96^3$
the above equation can be written as$ (100 + 4)^3 + (100 - 4)^3$
we know that,$ (a + b)^3 + (a - b)^3 = 2[a^3 + 3ab^2]$
here, $a= 100, b = 4$
$(100 + 4)^3 - (100 - 4)^{3 }$
$= 2[100^3 + 3(4)^2(100)]$
$= 2[1000000 + 4800]$
$= 2[1004800]$
$= 200960$
The value of $104^3 + 96^3 $
$= 2009600$
$104^3 + 96^3$
the above equation can be written as$ (100 + 4)^3 + (100 - 4)^3$
we know that,$ (a + b)^3 + (a - b)^3 = 2[a^3 + 3ab^2]$
here, $a= 100, b = 4$
$(100 + 4)^3 - (100 - 4)^{3 }$
$= 2[100^3 + 3(4)^2(100)]$
$= 2[1000000 + 4800]$
$= 2[1004800]$
$= 200960$
The value of $104^3 + 96^3 $
$= 2009600$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If O is the centre of the circle, find the value of x in the following figure:
Read the graph given in figure and answer the following question:
- What information is given by the bar given?
- In which years the areas under the sugarcane crop were the maximum and the minimum?
- State whether true or false:
A metallic sphere of radius $10.5\ cm$ is melted and then recast into smaller cones, each of radius $3.5\ cm$ and height $3\ cm.$ How many cones are obtained?
→In figure, ABCD and PQRC are rectangles and Q is the mid-point of AC. Prove that
→- DP = PC
- $\text{PR}=\Big(\frac{1}{2}\Big)\text{AC}$
Find the value $k$ if $x - 3$ is a factor of $k^2x^3 - kx^2 + 3kx - k.$
→In the given figure, m and it are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.
Prove that:
$\Big(\frac{1}{\text{x}^{\text{a}-\text{b}}}\Big)^{\frac{1}{\text{a}-\text{c}}}\times\Big(\frac{1}{\text{x}^{\text{b}-\text{c}}}\Big)^{\frac{1}{\text{b}-\text{a}}}\times\Big(\frac{1}{\text{x}^{\text{c}-\text{a}}}\Big)^{\frac{1}{\text{c}-\text{b}}}=1$
→$\Big(\frac{1}{\text{x}^{\text{a}-\text{b}}}\Big)^{\frac{1}{\text{a}-\text{c}}}\times\Big(\frac{1}{\text{x}^{\text{b}-\text{c}}}\Big)^{\frac{1}{\text{b}-\text{a}}}\times\Big(\frac{1}{\text{x}^{\text{c}-\text{a}}}\Big)^{\frac{1}{\text{c}-\text{b}}}=1$
It being given that $\sqrt{3}=1.732,\sqrt{5}=2.236,\sqrt{6}=2.449$ and $\sqrt{10}=3.162,$ find to three places of decimal, the value of the following:
$\frac{1}{4\sqrt{3}-3\sqrt{5}}$
→$\frac{1}{4\sqrt{3}-3\sqrt{5}}$
In figure AE bisects $\angle\text{CAD}$ and $\angle\text{B}=\angle\text{C}.$ Prove that AE || BC.
→Find the median of the data:
92, 35, 67, 85, 72, 81, 56, 51, 42, 69
92, 35, 67, 85, 72, 81, 56, 51, 42, 69