If a = 355, b = 356, c = 357, then find the value of $a^3+b^3+c^3-3abc=$

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If a = 355, b = 356, c = 357, then find the value of

a^3+b^3+c^3-3abc=

A3204

B3205

C3207

D3206

Answer:

The expression $a^3+b^3+c^3-3abc$ is a fundamental algebraic identity.
The general form of this identity is: $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ .
Another useful form, especially when dealing with differences between terms, is: $a^3+b^3+c^3-3abc = \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$ . This form highlights the squared differences, which are always non-negative.

When the numbers $a, b, c$ are in an Arithmetic Progression (AP), it means there is a constant difference between consecutive terms. Let this common difference be 'd'.
So, we can write $b = a+d$ and $c = a+2d$ . Alternatively, if 'b' is the middle term, then $a = b-d$ and $c = b+d$ .
For the special case where $a, b, c$ are in AP, the identity simplifies significantly to: $a^3+b^3+c^3-3abc = 3d^2(a+b+c)$ . This formula is extremely useful for competitive exams as it drastically reduces calculation time.
Derivation Hint: Substitute $a=b-d$ and $c=b+d$ into the general identity $\frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$ . Here, $(a-b) = -d$ , $(b-c) = -d$ , and $(c-a) = -2d$ . The sum $(a+b+c)$ also equals $3b$ .

Given $a = 355$ , $b = 356$ , $c = 357$ .
Observe that these numbers are consecutive integers, which means they are in an Arithmetic Progression.
The common difference (d) is $b-a = 356-355 = 1$ , or $c-b = 357-356 = 1$ . So, $d=1$ .
The sum $(a+b+c)$ is $355+356+357$ . Since it's an AP, the sum can also be calculated as $3 \times \text{middle term} = 3 \times b = 3 \times 356 = 1068$ .
Using the simplified formula for AP: $a^3+b^3+c^3-3abc = 3d^2(a+b+c)$ .
Substitute the values: $3 \times (1)^2 \times (1068) = 3 \times 1 \times 1068 = 3204$ .

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