here are the factorizations for the given expressions.
i) We need to factorize 125x3+y3.
This expression is in the form of a3+b3.
We use the identity a3+b3=(a+b)(a2−ab+b2).
Step 1: Identify a and b.
We can write 125x3 as (5x)3. So, a=5x.
And y3 is already in the form b3, so b=y.
Step 2: Substitute a and b into the identity.
125x3+y3=(5x+y)((5x)2−(5x)(y)+y2)
Step 3: Simplify the terms.
=(5x+y)(25x2−5xy+y2)
The factored form is:
(5x+y)(25x2−5xy+y2)
ii) We need to factorize 8x3−y3.
This expression is in the form of a3−b3.
We use the identity a3−b3=(a−b)(a2+ab+b2).
Step 1: Identify a and b.
We can write 8x3 as (2x)3. So, a=2x.
And y3 is already in the form b3, so b=y.
Step 2: Substitute a and b into the identity.
8x3−y3=(2x−y)((2x)2+(2x)(y)+y2)
Step 3: Simplify the terms.
=(2x−y)(4x2+2xy+y2)
The factored form is:
(2x−y)(4x2+2xy+y2)
iii) We need to factorize (2x−5y)3−(2x−y)3.
This expression is in the form of A3−B3.
We use the identity A3−B3=(A−B)(A2+AB+B2).
Step 1: Identify A and B.
Let A=(2x−5y) and B=(2x−y).
Step 2: Calculate A−B.
A−B=(2x−5y)−(2x−y)=2x−5y−2x+y=−4y
Step 3: Calculate A2, B2, and AB.
A2=(2x−5y)2=(2x)2−2(2x)(5y)+(5y)2=4x2−20xy+25y2
B2=(2x−y)2=(2x)2−2(2x)(y)+(y)2=4x2−4xy+y2
AB=(2x−5y)(2x−y)=2x(2x−y)−5y(2x−y)=4x2−2xy−10xy+5y2=4x2−12xy+5y2
Step 4: Calculate A2+AB+B2.
A2+AB+B2=(4x2−20xy+25y2)+(4x2−12xy+5y2)+(4x2−4xy+y2)
Combine like terms:
=(4x2+4x2+4x2)+(−20xy−12xy−4xy)+(25y2+5y2+y2)
=12x2−36xy+31y2
Step 5: Substitute the results from Step 2 and Step 4 into the identity A3−B3=(A−B)(A2+AB+B2).
(2x−5y)3−(2x−y)3=(−4y)(12x2−36xy+31y2)
Step 6: Distribute −4y into the trinomial.
=−4y⋅12x2−4y⋅(−36xy)−4y⋅31y2
=−48x2y+144xy2−124y3
The factored form is:
−48x2y+144xy2−124y3
Drop the next question!