factorise using identity 27a^3+ 1/64b^3+27a^2/4b+9a/16b^2
Answers
Answered by
0
Answer:
30192+$627#*&@!#&-;@#(+
Step-by-step explanation:
g ok I will never forget you
Answered by
0
Answer:
Consider the given expression
27a
3
+
64b
3
1
+
4b
27a
2
+
16b
2
9a
………..(1)
We know that (x+y)
3
=x
3
+y
3
+3x
2
y+3xy
2
+y
3
…………(2)
Thus,
27a
3
+
64.b
3
1
+
4b
27.a
2
+
16.b
2
9a
=(3a)
3
+
(4)
3
,b
3
1
+3(3a)
2
.(
4b
1
)+3(3a)(
4b
1
)
2
……….(3)
Comparing the above two equations (2) and (3), we have
x=3a;y=
4b
1
Thus equation (1) becomes.
27a
3
+
64b
3
1
+
4b
27.a
2
+
16.b
2
9a
=[3a+
4b
1
]
3
.
Similar questions