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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.
2x, 3x2, 4
Factorize the following:
20a12b2 − 15a8b4
Factorize the following:
10m3n2 + 15m4n − 20m2n3
Factorize the following:
ax2y + bxy2 + cxyz
Factorise:
`x^4 + y^4 - 27x^2y^2`
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Factorise : 9x 2 + 3x - 8y - 64y2
Factorise : `1/4 ( a + b )^2 - 9/16 ( 2a - b )^2`
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Factorise : 15x4y3 - 20x3y
Factorise : 17a6b8 - 34a4b6 + 51a2b4
Factorise : 3x5y - 27x4y2 + 12x3y3
factorise : 35a3b2c + 42ab2c2
factorise : x2y - xy2 + 5x - 5y
Factorise:
a2 – ab(1 – b) – b3
factorise : (ax + by)2 + (bx - ay)2
Factorise xy2 - xz2, Hence, find the value of :
9 x 82 - 9 x 22
Factorise the following by taking out the common factors:
12a3 + 15a2b - 21ab2
Factorise:
`"p"^2 + (1)/"p"^2 - 3`
