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# If a∗b = ${{a}^{3}}+{{b}^{3}}$ on Z, then ${{(2*3)}^{2}}$=(a) 925(b) 1625(c) 35(d) 1225

Last updated date: 17th Jul 2024
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Hint: For solving this problem, we should know the basics of given relations between two integers and use that relation to find the value. In this case we have a binary operator such that its operation on two integers (a and b) yields ${{a}^{3}}+{{b}^{3}}$. We use this to solve the above problem.

For solving a problem related to binary operations of relation concerning the one given in the problem, we should be aware about how binary operators work in general. We understand this through an example. Let’s assume a binary relation between a and b such that a*b = a+b. In this case, the binary operation between a and b states that we would perform the sum of a and b. Thus, for this operation, if we have to find 3*5, we would have the answer as 3+5 = 8. Now, coming back to the problem in hand. We have a different binary operation given by a∗b = ${{a}^{3}}+{{b}^{3}}$. Thus, we have,

= ${{(2*3)}^{2}}$

= ${{({{2}^{3}}+{{3}^{3}})}^{2}}$

= ${{(8+27)}^{2}}$

= ${{35}^{2}}$

= 1225

Hence, the correct answer is (d) 1225.

Note: In general, solving a problem related to binary operations is similar to decoding a thing based on the required constraints. For example, in the given problem, we were given a∗b = ${{a}^{3}}+{{b}^{3}}$. Thus, with this in mind, if given to evaluate c*d (for the domain on Z), we can decode this similar to the constraint, thus getting c∗d = ${{c}^{3}}+{{d}^{3}}$.
Also, note that we can use this property only for the integers (Z), since in the problem it is given that this property is only for integers. Thus, we cannot evaluate, say $\left( \dfrac{1}{3}*\dfrac{1}{2} \right)$ since, these are fractions.