Question

Factorise completely : 48x^4 – 243y^4​

Answer 1

Answer:

$(3)(4x^{2} +9y^{2})(2x+3y)(2x-3y)$

Step-by-step explanation:

We take out 3 common.

Then equation simplifies to $(3)(16x^{4}-81y^{4})$.

Take $x^{2} =a$ and $y^{2} =b$.

Then equation becomes  $(3)(16a^{2} -81b^{2})$.

Simplifying $(16a^{2} -81b^{2})$

$(16a^{2} -81b^{2})$=$(4a+9b)(4a-9b)$

=$(3)$$(4a+9b)(4a-9b)$

But $x^{2} =a$ and $y^{2} =b$

=$(3)(4x^{2}+9y^{2} )(4x^{2} -9y^{2} )$

But $(4x^{2} -9y^{2} )$=$(2x+3y)(2x-3y)$

Final answer = $(3)(4x^{2} +9y^{2})(2x+3y)(2x-3y)$.

Hope this helps. Good day to you.

Answer 2

Answer:

i can not getting your point this is lendi method very sad