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Class 9 Exam  >  Class 9 Questions  >  Factorise:(3x- 5y)3+ (5y – 2z)3+ (2z &n... Start Learning for Free
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Factorise:(3x- 5y)3+ (5y – 2z)3 + (2z – 3x)3

  • a)
    3(3x + 5y) (5y + 2z) (2z + 3x)

  • b)
    2(3x + 5y) (5y + 2z) (2z + 3x)

  • c)
    3(3x - 5y) (5y – 2z) (2z – 3x)

  • d)
    2(3x – 5y) (5y – 2z) (2z – 3x)

Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Factorise:(3x- 5y)3+ (5y – 2z)3+ (2z – 3x)3a)3(3x + 5y) (5...
The given expression is (3x - 5y)^3 (5y).

To factorize this expression, we can start by separating the expression inside the parentheses, (3x - 5y), and then factorize it using the binomial formula:

(3x - 5y)^3 (5y) = (3x - 5y)(3x - 5y)(3x - 5y)(5y)

Now, we can factor out the common factor (3x - 5y) from each term:

(3x - 5y)(3x - 5y)(3x - 5y)(5y) = (3x - 5y)^3 (3x - 5y)(5y)

Therefore, the factored form of the expression is (3x - 5y)^3 (3x - 5y)(5y).
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Factorise:(3x- 5y)3+ (5y – 2z)3+ (2z – 3x)3a)3(3x + 5y) (5...
(3x-5y)³+(5y-2z)³+(2z-3x)³ is the question. We don't have to cube them all to factorise. Now, One thing should be remembered that, in time of this type of factorisation we have to first add the polynomials to check if the sum is zero or not.If it is zero, then the factorisation will be in form of 3(abc). Now, lets sum them, (3x-5y) +(5y-2z) +(2z-3x).The answer is zero. So put the expression in form of 3(abc). So the factorisation of the expression is 3(3x-5y)(5y-2z)(2z-3x).ans
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Factorise:(3x- 5y)3+ (5y – 2z)3+ (2z – 3x)3a)3(3x + 5y) (5y + 2z) (2z + 3x)b)2(3x + 5y) (5y + 2z) (2z + 3x)c)3(3x - 5y) (5y – 2z) (2z – 3x)d)2(3x – 5y) (5y – 2z) (2z – 3x)Correct answer is option 'C'. Can you explain this answer?
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