please help me with this math..?

Question

What is the only positive value of b for which (x+b)^2 = x^2+2bx+b^3 is an identity for all real values of x?

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If x + y = 1, x^3+y^3 = 19, find the values of x^2 + y^2.


please show me the step by step process.

Answer 1

In the first equation, square the left member to get
x² + 2bx + b² = x² + 2bx + b^3 and subtract like terms from both sides leaving
b² = b^3 dividing both sides by b² gives b = 1

Looking at the next 2 equations, it's pretty obvious that one of the x or y must be negative and that their absolute difference must be 1. Now we need to find a pair of perfect cubes whose difference is 19. Fortunately, we don't have to think too hard since 2^3 is 8 and 3^3 is 27. So let x = 3 and y = -2 and
3+(-2) = 1 and
3^3 + (-2)^3 = 27 - 8 = 19 and 3² + (-2)² = 13.

Doug

Answer 2

(x+b)^2 = x^2 +2bx + b^3 (Given)
(x+b)^2 = x^2 +2bx + b^2 (Formula)
On comparing the two we get
b^3 = b^2
b = 0 or b = 1
Since b is positive, b =1

x^3 + y^3 =19
(x + y) (x^2 -xy +y^2) = 19
1(x^2 -xy +y^2) = 19
x^2 +y^2 = 19 + xy

(x + y) = 1
x^2 + 2xy +y^2 =1
x^2 +y^2 = 1 -2xy

19 + xy = 1 - 2xy
xy + 2xy = 1 - 19
3xy = -18
xy = -6

x + y = 1; y = 1 - x
Substituting y = 1 - x in xy = -6 we get
x(1-x) = -6
x - x^2 +6 = 0
x^2 -x -6 =0 (changing signs)
(x-3) (x+2) = 0
x =3 or x = -2

y = 1 - x = 1 - 3 = -2 or y = 1 - x = 1 - (-2) = 1 + 2 = 3
Hence x = 3, y = -2 or x = -2, y = 3
x^2 +y^2 = 9 + 4 = 13

Answer 3

the first one is quite simple.

ii) x^3 + y^3 =

19 = (x+y)^3 - 3xy(x+y)

19 = 1^3 + 3xy(1)

1 - 3xy = 19

3xy = -18
xy = -6 ...............(1)


x^3 + y^3 = (x+y)(x^2 - xy +y^2)

19 = 1(x^2 + y^2 - (-6)) { because we have proved earlier that
xy = 6}

19 = x^2 + y^2 + 6

x^2 + y^2 = 13.

Answer 4

1. (x+b)^2 = x^2 + 2 xb + b^2
now, to have b^2 = b^3, only possible sol'n is b=1
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2.
x + y = 1
so, (x+y)^3 = x^3+y^3 + 3xy(x+y) =1
substituting the values,
19+3xy(1)=1
so, xy= -6.
Now, (x+y)^2= x^2 + y^2 + 2 xy = 1
so, x^2+y^2 + 2 (-6) = 1
so, x^2+y^2= 13

Answer 5

(x+b)^2 = x^2+2bx+b^3
x^2+2bx+b^2 = x^2+2bx+b^3
--> expand left hand side
b^2 = b^3
--> subtract x^2 and 2bx from both sides
b = 1
--> divide both sides by b^2
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i don't know the 2nd one

Answer 6

I would tell you this in great detail, but judging by your name, you're a man hater, so I think I'll not give you help on your homework.

Answer 7

i have already answer your first question posted a while ago...