Cube = Square - Square

We have this pattern:

• 1³ = 1² - 0²
• 2³ = 3² - 1²
• 3³ = 6² - 3²
• 4³ = 10² - 6²
• 5³ = x² - 10²
• 6³ = y² - x²
• 7³ = z² - y²

So...

• x² = 5³ + 10² = 225 ∴ x = 15 (We ignore, always, the negative number. In this case, -15.)
• y² = 6³ + 15² = 441 ∴ y = 21
• z² = 7³ + 21² = 784∴ y = 28

If a = 21 and b = 28, 28 * 21 = 558.

First notice the pattern. Starting with the 2nd equation, you can see that the $b^2$ component is the $a^2$ component from the previous equation. Furthermore, the difference between a and b increases by 1 each time, and, in fact, it is equal the base of the cube root.

$1^3 = 1^2 - 0^2$ : a - b = 1 = base, $2^3 = 3^2 - 1^2$ : a - b = 2 = base, $3^3 = 6^2 - 3^2$ : a - b = 3 = base, ....ect.

Following this pattern we eventually come to

$7^3 = 28^2 - 21^2$ and therefore $a \times b = 588$

The pattern is easy to discover. Each $k$ th cube can be expressed by letting $m^2$ be the last term of the previous expression: $k^3 = (m+k)^2 - m^2$ . Extrapolating the pattern forwards we get: $5^3 = 15^2 - 10^2, 6^3 = 21^2 - 15^2, 7^3 = 28^2 - 21^2$

$28\times 21 = \boxed{588}$ which is our answer.

We prove the validity of this pattern by induction. Several bases have already been established in the question. Let us assume for some $k$ , $k^3 = m^2 - (m - k)^2 = (2m - k)(k) \implies 2m - k = k^2$ .

Adding $2k + 1$ to both sides we get:

$2m + k + 1 = (k + 1)^2 \implies (2m + k + 1)(k+1) = (k + 1)^3 = (m + k + 1)^2 - m^2$

This completes the proof.

From the pattern, we notice for $n^{3}$ , where n is a non-negative integer,

$n^{3}$ = $a^{2}$ - $b^{2}$ , where a and b are non-negative integers.

From the pattern, a = n + b. So, for the specific case where n = 7,

$7^{3}$ = $(7+b)^{2}$ - $b^{2}$

We manipulate this equation and solve for the unknown 'b' by expanding the right hand side of the equation.

343 = 49 + 14b + $b^{2}$ - $b^{2}$

294 = 14b

b = 21

a = 7 + b

a = 7 + 21

a = 28

a x b = 28 x 21

a x b = 588

∴ The answer is 588.

The cube of the nth number is equal to the difference of the square of the nth triangular number and the square of the (n-1)th triangular number. Hence, 7 cubed equals 28 squared minus 21 squared. We can also write this shorthand as $7^3 = 28^2-21^2$ . Therefore $a$ and $b$ are 28 and 21 respectively, and $28 * 21 = 588$

let: $z^3 = x^2 - y^2$ then $z^3 = (x+y)(x-y) \Rightarrow z = \frac{(x+y)}{(x-y)}$ ,

and let $y = z*\alpha$ and $x = y + z =z + z * \alpha$

$z = \frac{(x+y)}{(x-y)} = \frac{(z + z * \alpha + z * \alpha)}{(z + z * \alpha - z * \alpha)}$

$z = \frac{(z + 2 * z * \alpha )}{z}$

$\alpha = \frac{(z^2 - z)}{(2 * z)}$

then for $7$ we've:

$\alpha = \frac{(7^2 - 7)}{(2 * 7)} = 3$

$x = y + 7 , y = \alpha * 7$

$x = 28 , y = 21$

$x * y = 28 * 21 = 588$

$c_{1}^{3}=(c_{2})^{2}-(c_{2}-c_{1})^{2}$ , $c_{1}^{3}=(\frac{c_{1}(c_{1}+1)}{2})^{2}-(c_{2}-c_{1})^{2}$ , $7^{3}=(\frac{(7)(8)}{2})^{2}-(\frac{(7)(8)}{2})-7)^{2}$ , $7^{3}=(28)^{2}-(21)^{2}$ , $a=28, b=21$ $a \times b=28 \times 21=588$

7^3=(A-B)(A+B) A-B=7 A+B=49 2A=56 A=28 B=21 AB=28*21=588

the solution is: a= +2 +3 +4.... if 7 so that +7 that is 28 b= +1+2+3...... if 7 so that + 6 that is 21 28 * 21 = 588

1-3=2 3-6=3 6-10=-4 so next terms are 15 '21'28 by this 7cube is 28squre-21squre

using calculator f(x) es plus 991 - 7^3=x^2- a^2 the a in - 6^3= A^2 - b ^2= 21 the b is in - 5^3=10 which we got from 4^3 using that 7^3=x^2- 21^2 ans= 28 28 21=588

7^3=(a+b)(a-b)...a+b=7^2=49, a-b=7 solving we get a=28 b=21 axb=588