Question 24 of 43beginner💡 UnderstandShort Answer2 marks

Verify this for yourself by choosing different values of x and y. (Context: Thus (x - y)(x2x^{2} + xy + y2y^{2}) = x3x^{3} - y3y^{3}. This is also an identity!)

Correct Answer

To verify the identity (x−y)(x2+xy+y2)=x3−y3(x - y)(x^2 + xy + y^2) = x^3 - y^3, let us choose different values for xx and yy. First, let x=3x = 3 and y=1y = 1. Substituting these values, the LHS becomes (3−1)(32+3(1)+12)=2×(9+3+1)=2×13=26(3 - 1)(3^2 + 3(1) + 1^2) = 2 \times (9 + 3 + 1) = 2 \times 13 = 26. The RHS becomes 33−13=27−1=263^3 - 1^3 = 27 - 1 = 26. Since LHS = RHS, the identity holds true.

Next, let x=4x = 4 and y=2y = 2. Substituting these, the LHS is (4−2)(42+4(2)+22)=2×(16+8+4)=2×28=56(4 - 2)(4^2 + 4(2) + 2^2) = 2 \times (16 + 8 + 4) = 2 \times 28 = 56. The RHS is 43−23=64−8=564^3 - 2^3 = 64 - 8 = 56. Again, LHS = RHS. Thus, the identity is verified for different values of xx and yy.

Exercise: In-text Verification | Q: 1 | (Page 18)
For More Understanding

Explanation

The textbook encourages students to verify the identity by substituting different values for the variables. By choosing simple positive integers, students can easily calculate both the LHS and RHS to see that they match, reinforcing the concept that an identity is true for all values of the variables.

Solution Steps

  1. Step 1: Choose a set of values for x and y, e.g., x = 3, y = 1.

  2. Step 2: Calculate LHS: (3 - 1)(323^{2} + 3(1) + 121^{2}) = 2 * 13 = 26.

  3. Step 3: Calculate RHS: 333^{3} - 131^{3} = 27 - 1 = 26. LHS = RHS.

  4. Step 4: Choose another set of values, e.g., x = 4, y = 2.

  5. Step 5: Calculate LHS: (4 - 2)(424^{2} + 4(2) + 222^{2}) = 2 * 28 = 56.

  6. Step 6: Calculate RHS: 434^{3} - 232^{3} = 64 - 8 = 56. LHS = RHS. Identity verified.

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