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without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³
Most Upvoted Answer
without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³
By identity if a+b+c=0 then a^3+ b^3+ c^3=3abc
x-2y+2y-3z+3z-x=0
0=0
then (x-2y)^3+(2y-3z)^3+(3z-x)^3=3(x-2y)(2y-3z)(3z-x)
x-2y+2y-3z+3z-x=0
0=0
then (x-2y)^3+(2y-3z)^3+(3z-x)^3=3(x-2y)(2y-3z)(3z-x)
Community Answer
without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³
Factorization of (x-2y)³ + (2y-3z)³ + (3z-x)³
To factorize the given expression without finding cubes, we can use the formula for sum of cubes, which states that \(a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc)\).
Here, we have:
a = x, b = -2y, c = 2y
d = 2y, e = -3z, f = 3z
g = 3z, h = -x, i = -2y
Applying the formula
Now, substituting these values into the formula, we get:
\((x - 2y)^3 + (2y - 3z)^3 + (3z - x)^3 = (x - 2y + 2y - 3z + 3z - x)((x - 2y)^2 + (2y - 3z)^2 + (3z - x)^2 - (x - 2y)(2y - 3z) - (x - 2y)(3z - x) - (2y - 3z)(3z - x))\).
Simplifying the expression
Simplifying further, we get:
\((x - 2y)^3 + (2y - 3z)^3 + (3z - x)^3 = (0)((x - 2y)^2 + (2y - 3z)^2 + (3z - x)^2 - (x - 2y)(2y - 3z) - (x - 2y)(3z - x) - (2y - 3z)(3z - x))\).
Therefore, the factorized form of the given expression is 0.
To factorize the given expression without finding cubes, we can use the formula for sum of cubes, which states that \(a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc)\).
Here, we have:
a = x, b = -2y, c = 2y
d = 2y, e = -3z, f = 3z
g = 3z, h = -x, i = -2y
Applying the formula
Now, substituting these values into the formula, we get:
\((x - 2y)^3 + (2y - 3z)^3 + (3z - x)^3 = (x - 2y + 2y - 3z + 3z - x)((x - 2y)^2 + (2y - 3z)^2 + (3z - x)^2 - (x - 2y)(2y - 3z) - (x - 2y)(3z - x) - (2y - 3z)(3z - x))\).
Simplifying the expression
Simplifying further, we get:
\((x - 2y)^3 + (2y - 3z)^3 + (3z - x)^3 = (0)((x - 2y)^2 + (2y - 3z)^2 + (3z - x)^2 - (x - 2y)(2y - 3z) - (x - 2y)(3z - x) - (2y - 3z)(3z - x))\).
Therefore, the factorized form of the given expression is 0.
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without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³
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without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³ for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³ covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³.
without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³ for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³ covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for without finding cubes factorise (x-2y)³ +(2y-3z)³ +(3z-x)³.
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