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प्रश्न
Factorise : 12abc - 6a2b2c2 + 3a3b3c3
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उत्तर
12abc - 6a2b2c2 + 3a3b3c3 = 3abc (4 - 2abc + a2b2c2)
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संबंधित प्रश्न
Find the common factors of the terms.
3x2y3, 10x3y2, 6x2y2z
Factorise the following expression:
10a2 − 15b2 + 20c2
Factorise the following expression:
ax2y + bxy2 + cxyz
Factorise by taking out the common factors :
2 (2x - 5y) (3x + 4y) - 6 (2x - 5y) (x - y)
Factorise:
`x^4 + y^4 - 27x^2y^2`
Factorise : `x^2 + 1/x^2 - 3`
Factorise : 2(ab + cd) - a2 - b2 + c2 + d2
Factorise : 15x + 5
Factorise : 4a2 - 8ab
Factorise : 15x4y3 - 20x3y
Factorise : a3b - a2b2 - b3
Factorise : (x + y)(a + b) + (x - y)(a + b)
factorise : x2y - xy2 + 5x - 5y
Factorise xy2 - xz2, Hence, find the value of :
9 x 82 - 9 x 22
Factorise: a4 - 625
Factorise: x4 - 5x2 - 36
Factorise the following by taking out the common factors:
12a3 + 15a2b - 21ab2
Factorise the following by taking out the common factors:
p(p2 + q2 - r2) + q(r2 - q2 -p2) - r(p2 + q2 - r2)
Factorise:
`"p"^2 + (1)/"p"^2 - 3`
