Week 2

Example 1. Find the value of 2021³ – 4043³ + 2022³.

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Solution 1. If we put the given values in a calculator, it will show us as below:

Of course, this is the correct answer. What to do when a calculator is not available? Looking at the given expression, following probable questions may come to our mind:

(i) Is it possible to apply the formula of a³ + b³ or a³ – b³ or any other?

(ii) Can we explore any kind of symmetry?

(iii) Do any rearrangement helps? Particularly, writing any term/ the expression in any specific form?

Yes, here writing the term(s) in specific manner will help us. Recall that (–a)³ = –a³ and hence –4043³ = (–4043)³. This is the remarkable step to achieve the solution.

Now our question is to find the value of 2021³ + (–4043)³ + 2022³. Anything special?

Yes. Observe that 2021 + (–4043) + 2022 = 0. Therefore, we may consider the following theorem:

Theorem. If a + b + c = 0 then a³ + b³ + c³ = 3abc.

Since 2021 + (–4043) + 2022 = 0, applying the above theorem, we have

2021³ + (–4043)³ + 2022³ = 3(2021)(–4043)(2022) = –49564697598.

Instead, if we want to calculate it directly using the formula for a³ + b³ and a³ – b³, we get

2021³ – 4043³ + 2022³ = (2021³ – 4043³) + 2022³

= (2021 – 4043)(2021² + 2021 × 4043 + 4043²) + 2022³

= (–2022)(2021² + 2021 × 4043 + 4043²) + 2022³

= (–2022)(2021² + 2021 × 4043 + 4043² – 2022²)

= (–2022)[(2021² – 2022²) + 2021 × 4043 + 4043²)]

= (–2022)[(2021 + 2022) (2021 – 2022) + 2021 × 4043 + 4043²]

= (–2022)[(4043)(–1) + 2021 × 4043 + 4043²]

= (–2022)(4043)[–1 + 2021 + 4043]

= (–2022)(4043)(6063) = (–2022)(4043)(3 × 2021)

= 3(2022)(–4043)(2021)

= –49564697598, as before.

Solution 2. Recall that, any two-digit number say ab, can always be written as 10a + b. For example, 26 = 10 × 2 + 6; 65 = 10 × 6 + 5.

Rearranging the equation, we get 9xz = y(10x – z) and we need to solve this equation for x, y and z.

Now, y must be 3 or 6 or 9. If not, then (10x – z)must be divisible by 9. But this is possible only when x = z. Under this condition, from the symmetry of the given question, we have

x(10x + y) = x(10y + x)    or   10x² + xy = 10xy + x²     or     x² = xy.     

We conclude that, either x = 0 or x = y. Of course, x = 0 is not possible and x = y will ultimately lead to x = y = z, an obvious impossibility.

Case 1. Let y = 3. Then from 9xz = y(10x – z), we get 3xz = 10x – z.

If x = 1, then 3z = 10 – z and no positive integer solution is possible.

In a similar way, there will be no positive integer solution if x = 2, 3, . . . , 9.

Case 2. Let y = 6. Then from 9xz = y(10x – z), we get 3xz = 2(10x – z).

If x = 1, then 3z = 20 – 2z and z = 4 is a solution.

If x = 2, then 3z = 20 – z and z = 5 is a solution.

If x = 3, then 9z = 2(30 – z) and no positive integer solution is possible.

In a similar way, there will be no positive integer solution if x = 4, 5, . . . , 9.

Case 3. Let y = 9. Then from 9xz = y(10x – z), we get xz = (10x – z).

If x = 1, then z = 10 – z and z = 5 is a solution.

If x = 4, then 4z = 40 – z and z = 8 is a solution.

If x = 2, then 2z = 20 – z and no positive integer solution is possible.

In a similar way, there will be no positive integer solution if x = 3, 5, . . . , 9.

Thus, we obtain (x, y, z) = (1, 6, 4), (2, 6, 5), (1, 9, 5) and (4, 9, 8).