Consider the following statements :
`1.` If `A={(x,y) in [RxxR : x^(3)+y^(3)=1]}` and `B={(x,y) in [R: (x-y)=1]}`, then `AnnB` contains exactly one elements.
`2.` If `A={(x,y) in [RxxR : x^(3)+y^(3)=1]}` and `B={(x,y) in [R: (x+y)=1]}`, then `AnnB` contains exactly two elements.
which of the above statements is/are correct ?

Options

A. `1` only
B. `2` only
C. Both `1` and `2`
D. Neither `1` nor `2`

Answer

To solve the problem, we will analyze both statements one by one.

### Statement 1:
**Given:**
- Set \( A = \{(x,y) \in \mathbb{R} \times \mathbb{R} : x^3 + y^3 = 1\} \)
- Set \( B = \{(x,y) \in \mathbb{R} \times \mathbb{R} : x - y = 1\} \)

**Step 1:** Substitute \( x \) from set \( B \) into set \( A \).

From \( B \), we have:
\[ x = y + 1 \]

**Step 2:** Substitute \( x \) into the equation of set \( A \).

Substituting into \( A \):
\[ (y + 1)^3 + y^3 = 1 \]

**Step 3:** Expand the left-hand side.

Expanding \( (y + 1)^3 \):
\[ y^3 + 3y^2 + 3y + 1 + y^3 = 1 \]
This simplifies to:
\[ 2y^3 + 3y^2 + 3y + 1 = 1 \]

**Step 4:** Simplify the equation.

Subtracting 1 from both sides:
\[ 2y^3 + 3y^2 + 3y = 0 \]

**Step 5:** Factor out \( y \).

Factoring out \( y \):
\[ y(2y^2 + 3y + 3) = 0 \]

**Step 6:** Solve for \( y \).

This gives us one solution:
\[ y = 0 \]
Now, we need to check the quadratic \( 2y^2 + 3y + 3 = 0 \) for real solutions.

**Step 7:** Calculate the discriminant.

The discriminant \( D \) is given by:
\[ D = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 3 = 9 - 24 = -15 \]
Since \( D < 0 \), there are no real solutions from this quadratic.

**Step 8:** Find \( x \) corresponding to \( y = 0 \).

Substituting \( y = 0 \) back into \( x = y + 1 \):
\[ x = 0 + 1 = 1 \]

Thus, the only solution in \( A \cap B \) is:
\[ (1, 0) \]

### Conclusion for Statement 1:
**Statement 1 is true.** \( A \cap B \) contains exactly one element: \( (1, 0) \).

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### Statement 2:
**Given:**
- Set \( A = \{(x,y) \in \mathbb{R} \times \mathbb{R} : x^3 + y^3 = 1\} \)
- Set \( B = \{(x,y) \in \mathbb{R} \times \mathbb{R} : x + y = 1\} \)

**Step 1:** Substitute \( x \) from set \( B \) into set \( A \).

From \( B \):
\[ x = 1 - y \]

**Step 2:** Substitute \( x \) into the equation of set \( A \).

Substituting into \( A \):
\[ (1 - y)^3 + y^3 = 1 \]

**Step 3:** Expand the left-hand side.

Expanding \( (1 - y)^3 \):
\[ 1 - 3y + 3y^2 - y^3 + y^3 = 1 \]
This simplifies to:
\[ 1 - 3y + 3y^2 = 1 \]

**Step 4:** Simplify the equation.

Subtracting 1 from both sides:
\[ -3y + 3y^2 = 0 \]

**Step 5:** Factor out \( 3y \).

Factoring out \( 3y \):
\[ 3y(y - 1) = 0 \]

**Step 6:** Solve for \( y \).

This gives us two solutions:
\[ y = 0 \quad \text{and} \quad y = 1 \]

**Step 7:** Find \( x \) corresponding to each \( y \).

For \( y = 0 \):
\[ x = 1 - 0 = 1 \quad \Rightarrow \quad (1, 0) \]

For \( y = 1 \):
\[ x = 1 - 1 = 0 \quad \Rightarrow \quad (0, 1) \]

### Conclusion for Statement 2:
**Statement 2 is true.** \( A \cap B \) contains exactly two elements: \( (1, 0) \) and \( (0, 1) \).

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### Final Conclusion:
Both statements are correct.

**Answer:** Both statements 1 and 2 are correct.

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