algebra - quadratics. pls help!! thank you

(a) We can use the difference of cubes factorization to solve for ab:

(a−b)(a2+ab+b2)=52

Since a−b=4, we can substitute this value to get:

4(a2+ab+b2)=52

Solving for a2+ab+b2, we get:

a2+ab+b2=13

We can then use the quadratic formula to solve for ab:

ab=2a−b±b2−4ac​​

Substituting a2+ab+b2=13 and a−b=4 into the quadratic formula, we get:

ab=2⋅4−4±42−4⋅13⋅4​​

ab=8−4±−112​​

ab=8−4±4isqrt(7)

Therefore, the possible values of ab are −2isqrt(7),2isqrt(7)​,−4,4​.

 

(b) We can use the sum of cubes factorization to solve for a+b:

(a+b)(a2−ab+b2)=a3+b3

Since a3−b3=52, we can substitute this value to get:

(a+b)(a2−ab+b2)=52

Solving for a+b, we get:

a+b=a2−ab+b2a3−b3​=1352​=4​

 

(c) Since we already found that ab can take on the values −2i7sqrt(7),2isqrt(7)​,−4,4, and a+b=4​, we can find all possible values of a and b by substituting these values into the equation a−b=4.

For ab=−2i7​, we get:

a−(−2i7​)=4

a=4−2i7​

For ab=2i7​, we get:

a−(2i7​)=4

a=4+2i7​

For ab=−4, we get:

a−(−4)=4

a=8

For ab=4, we get:

a−4=4

a=8

Therefore, the possible values of a and b are (4−2isqrt(7),8),(4+2isqrt(7)​,8),(8,8)​.

 a - b=4,     

a^3 - b^3=52, solve for a,  b

Use substitution to get:

 

a = 2 - sqrt(3) and  b = -2 - sqrt(3)

 

                       OR:

 

a = 2 + sqrt(3) and b = sqrt(3) - 2