Simplify the following expression:
33×31×(2
10
+1)
(3
2
)
4
−1
​

Let's simplify the numerator first:
Numerator=(3
2
)
4
−1=3
8
−1
We can rewrite 3
8
−1 using the difference of squares formula a
2
−b
2
=(a−b)(a+b):

3
8
−1=(3
4
)
2
−1
2
=(3
4
−1)(3
4
+1)
3
4
=81⟹3
4
−1=80 and 3
4
+1=82
So, Numerator=80×82

Now let's simplify the denominator:

Denominator=33×31×(2
10
+1)
2
10
=1024⟹2
10
+1=1025
So, Denominator=33×31×1025

Let's re-examine the structural values to find a cancellation pattern:
Notice that 3
8
−1=(3
4
−1)(3
4
+1)=(3
2
−1)(3
2
+1)(3
4
+1)=(9−1)(9+1)(81+1)=8×10×82=6560.

Let's re-verify the terms in the original question prompt:
The question shows:
33×31×(2
10
+1)
((3
2
)
4
−1)
​
which can sometimes have alternative typographic interpretations in standard Indian curriculum memory-based papers where (2
10
+1) might actually be a misprint for something else, or let's re-calculate:
Numerator=3
8
−1=6561−1=6560.

Let's calculate the value of the denominator expression directly from the prompt:
33×31=1023.
Notice that 1023=2
10
−1.
Therefore, the denominator is (2
10
−1)(2
10
+1)=(2
10
)
2
−1=2
20
−1.
But wait, let's match the bases or examine if there is an alternative representation:
33×31=1023=2
10
−1.
Thus, Denominator=(2
10
−1)(2
10
+1)=2
20
−1.
This does not easily simplify with 6560.

Let's look closely at the typographical structure of the question where powers might be shifted:
If the denominator is actually 33×31×something or if the numbers are related to base 2 or 3:
Let's check 3
8
−1=6560.
If the denominator is 33×31×… wait, 33×31=1023.
Let's look at another common question template:
33×31
(2
2
)
4
−1
​
? No, if base is 2: (2
2
)
4
−1=2
8
−1=255.
What if the original expression was
…
(3
2
)
4
−1
​
? Let's check 3
8
−1=6560.
Let's check 6560/80=82.

Let's check if the denominator contains a typo in the original test sheet where (2
10
+1) was written instead of something else, or if the calculation yields a simple integer like 2, 1, −2, −1.
Let's check if 3
8
−1 is a typo for (2
5
)
4
−1=2
20
−1.
If the numerator is (2
5
)
4
−1=2
20
−1, then:

Denominator=33×31×(2
10
+1)=(2
10
−1)(2
10
+1)=2
20
−1
In that case,
2
20
−1
2
20
−1
​
=1.
The expression in the prompt writes ((32)4-1) which is a classic OCR/typing error for ((2
5
)
4
−1) or (2
5
)
4
−1 because 2
5
=32!
Yes! (32)
4
−1=(2
5
)
4
−1=2
20
−1.
Let's substitute 32=2
5
:

Numerator=(32)
4
−1=(2
5
)
4
−1=2
20
−1
Denominator=33×31×(2
10
+1)=(2
5
+1)(2
5
−1)(2
10
+1)=(2
10
−1)(2
10
+1)=2
20
−1
Thus, the value of the expression is:

2
20
−1
2
20
−1
​
=1