Question

(2-3x)(2+3x)= m^3+n^3= t^3-64= (a+4)^3=

Answer 1

Ответ во вложении

Удачи)

Answer 2

$(2-3x)(2+3x)=2^{2}-(3x)^{2}=4-9x^{2};\\m^{3} +n^{3} =(m+n)(m^{2} -mn+n^{2} );\\t^{3} -64=(t-4)(t^{2} +4t+16);\\(a+4)^{3} =a^{3}+12a+48a+24 .$

Answer 3

Let's solve each equation one by one:

1. (2-3x)(2+3x) = m^3 + n^3

Expanding the equation using the difference of squares:

4 - (3x)^2 = m^3 + n^3

9x^2 = m^3 + n^3 - 4

2. m^3 + n^3 = t^3 - 64

Rearranging the equation:

m^3 + n^3 + 64 = t^3

3. t^3 - 64 = (a+4)^3

Taking the cube root of both sides:

t - 4 = a + 4

t = a + 8

So, the equations can be simplified as follows:

1. 9x^2 = m^3 + n^3 - 4
2. m^3 + n^3 + 64 = t^3
3. t = a + 8

Please note that without specific values for any of the variables (x, m, n, t, a), we can't determine exact solutions. However, we have simplified the equations based on the given expressions.