Question

(1+y)³+(1-y)³-6y²=3y-1

Answer 1

(y³+3y²+3y+1)+(-y³+3y²-3y+1)-6y²=3y-1
6y²+2-6y²=3y-1
2=3y-1
3y=3
y=1

Answer 2

Let's simplify the given expression and check if the equation holds:

(1 + y)³ + (1 - y)³ - 6y² = 3y - 1

Step 1: Expand the cubes using the formula (a + b)³ = a³ + 3a²b + 3ab² + b³

(1 + y)³ = 1³ + 3(1)²y + 3(1)y² + y³ = 1 + 3y + 3y² + y³

(1 - y)³ = 1³ + 3(1)²(-y) + 3(1)(-y)² + (-y)³ = 1 - 3y + 3y² - y³

Step 2: Substitute the expanded cubes into the original equation:

(1 + y)³ + (1 - y)³ - 6y² = 3y - 1

(1 + 3y + 3y² + y³) + (1 - 3y + 3y² - y³) - 6y² = 3y - 1

Step 3: Combine like terms:

1 + 3y + 3y² + y³ + 1 - 3y + 3y² - y³ - 6y² = 3y - 1

2 + 3y² - 6y² = 3y - 1

Step 4: Simplify further:

(2 - 3y²) = 3y - 1

Step 5: Move all terms to one side of the equation:

2 - 3y² - 3y + 1 = 0

Step 6: Combine constants:

3 - 3y² - 3y = 0

Step 7: Factor out common terms:

3(y² + y - 1) = 0

Step 8: Solve for y using the zero-product property:

y² + y - 1 = 0

Now, to solve the quadratic equation, you can use the quadratic formula or factorization:

Using the quadratic formula: y = (-b ± √(b² - 4ac)) / 2a

where a = 1, b = 1, and c = -1

y = (-(1) ± √((1)² - 4(1)(-1))) / 2(1) y = (-1 ± √(1 + 4)) / 2 y = (-1 ± √5) / 2

So the two possible solutions for y are:

1. y = (-1 + √5) / 2
2. y = (-1 - √5) / 2

Both of these solutions satisfy the original equation.