If a+b=1 and ... find a squared plus b squared. (full working to be shown)

Question

Q1) If a+b=1 and a³+b³=4, find, without working the values of a and b, the value of a2+b2. (working must be shown) 

Answer 1

a³ +b³ + a + b = 4 + 1 = 5

 

(a + b)(a² - ab + b²) + (a +b) = 5

 

1(a² - ab + b²) + 1 = 5

 

a² - ab +b² = 4   NOTE: b = (1 - a)

 

a² - a(1-a) +(1 - a)² = 4

 

a² - a + a² + 1 - 2a + a² = 4

 

3a² - 3a - 3 = 0 or a² - a - 1 = 0 

 

Similarly, if we set a = (b - 1), we get b² - b - 1= 0

 

a² - a - 1 + b² - b -1 = 0

 

a² + b² - (a + b) - 2 = 0

 

a² + b² - 1 - 2 = 0

 

a² + b² = 3

 

A²

 

 

 

 

 

 

 

Answer 2

Assuming the problem seeks to solve a2 in terms of b and b2 in terms of a then:

a2∗a+b3=4

solve for a

a+b=1 then a=1-b

substitute

a2(1-b)+b3=4

rewrite

a2(1-b)=4-b3

divide both sides by (1-b)

a2=(4-b3)/(1-b) 

To solve for b2 solve a+b=1 for a and substitute into a3+b2∗b=4

b2=(4-a3)/(1-a)

best I can do with the info provided in the problem!!

Can you message me to let me know that I have or have not interpreted the problem correctly? 

 

Answer 3

(a+b)³ = a³ + 3a²b + 3ab² + b³

 

           = 3a²b + 3ab² + 4  (since a³+b³ = 4)

 

           = 3ab(a+b) + 4

 

           = 3ab+4 (since a+b = 1)

 

But (a+b)³ = 1³ = 1

 

So, 3ab+4 = 1     ab = -1

 

(a+b)² = a² + 2ab + b² = a²+b²+2(-1)

 

                                    = a²+b²-2

 

But, (a+b)² = 1²=1

 

So, a²+b²-2 = 1    a²+b² = 3