factorise (27x^3+27x^2+9x+1)
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Answer:
(3x+l)^3
Step-by-step explanation:
(3x)^3+(1)^3+3×3x^2×1+3×3x1^2
(3x+1)^3
(3x+1) (3x+1) (3x+1)
Answer:
(3x+1)(3x+1)(3x+1)
Step-by-step explanation:
27x^3 + 27x^2 + 9x + 1 ........(1)
Firstly check whether 3x+1 is factor of the given polynomial or not
Let 3x + 1 = 0
3x = -1
x = -1/3
Put the value of x in Eq.(1)
=> 27(-1/3)^3 + 27(-1/3)^2 + 9(-1/3) + 1
=> 27(-1/27) + 27(1/9) + 9(-1/3) + 1
=> -1 + 3 - 3 + 1
=> 0
Therefore, it is factor of the given polynomial
On dividing 27x^3 + 27x^2 + 9x + 1 by 3x +1, we get
9x^2 + 6x + 1
(for division see the attachment)
Now, factorisation of 9x^2 + 6x + 1
=> 9x^2 + 6x + 1
=> 9x^2 + 3x + 3x +1
=> 3x(3x + 1) + 1(3x + 1)
=> (3x + 1)(3x +1)
Therefore, factorisation of 27x^3 +;27x^2 + 9x + 1 is
=> (3x + 1)(3x + 1)(3x + 1)
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