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(i) 8x^3 – y^3 = (2x – y) (4x^2 + 2xy + y^2) (ii) 16x^2 – 9y^2 = (4x – 3y) (4x + 3y) - Mathematics

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Question

(i) 8x3 – y3 = (2x – y) (4x2 + 2xy + y2)

(ii) 16x2 – 9y2 = (4x – 3y) (4x + 3y)

Options

  • Only (i)

  • Only (ii)

  • Both (i) and (ii)

  • Neither (i) nor (ii)

MCQ
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Solution

Both (i) and (ii)

Explanation:

Let’s analyze each:

(i) 8x3 – y3:

We can rewrite (8x3) as (2x)3, so the expression becomes:

(2x)3 – y3

This fits the difference of cubes factorization formula:

a3 – b3 = (a – b)(a2 + ab + b2)

Applying it here:

(2x – y)((2x)2 + (2x)(y) + y2) = (2x – y)(4x2 + 2xy + y2)

Which matches the given factorization for (i).

(ii) 16x2 – 9y2:

This expression is a difference of squares, since:

16x2 = (4x)2

9y2 = (3y)2

So applying the difference of squares formula:

a2 – b2 = (a – b)(a + b)

We get:

(4x – 3y)(4x + 3y)

which matches the given statement (ii).

Therefore, both (i) and (ii) are correctly factorized.

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Chapter 4: Factorisation - Exercise 4F [Page 91]

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Nootan Mathematics [English] Class 9 ICSE
Chapter 4 Factorisation
Exercise 4F | Q 2. | Page 91
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