observe the picture and write a paragraph on it
Answers
Answer:
Hint.
Let A=2a,B=3b,C=4cA=2a,B=3b,C=4c . We observe that A^2=4a^2,B^2=9b^2,C^2=16c^2A
2
=4a
2
,B
2
=9b
2
,C
2
=16c
2
.
Also, we can observe that the sum of the bases of three cubes equal to zero. We have a suitable identity a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca)a
3
+b
3
+c
3
−3abc=(a+b+c)(a
2
+b
2
+c
2
+ab+bc+ca) since it is a sum of three cubes. If the sum of bases are equal to zero, one of the factors is zero, a+b+c=0a+b+c=0 . Consequently a^3+b^3+c^3=3abca
3
+b
3
+c
3
=3abc .
Solution.
Given, \dfrac{(4a^2-9b^2)^3+(9b^2-16c^2)^3+(16c^2-4a^2)^3}{(2a-3b)^3+(3b-4c)^3+(4c-2a)^3}
(2a−3b)
3
+(3b−4c)
3
+(4c−2a)
3
(4a
2
−9b
2
)
3
+(9b
2
−16c
2
)
3
+(16c
2
−4a
2
)
3
=\dfrac{(A^2-B^2)+(B^2-C^2)^3+(C^2-A^2)^3}{(A-B)^3+(B-C)^3+(C-A)^3}=
(A−B)
3
+(B−C)
3
+(C−A)
3
(A
2
−B
2
)+(B
2
−C
2
)
3
+(C
2
−A
2
)
3
=\dfrac{3(A^2-B^2)(B^2-C^2)(C^2-A^2)}{3(A-B)(B-C)(C-A)}=
3(A−B)(B−C)(C−A)
3(A
2
−B
2
)(B
2
−C
2
)(C
2
−A
2
)
=(A+B)(B+C)(C+A)=(A+B)(B+C)(C+A)
Substituting A=2a,B=3b,C=4cA=2a,B=3b,C=4c back into the equation,
=(2a+3b)(3b+4c)(4c+2a)=(2a+3b)(3b+4c)(4c+2a)
=\boxed{2(2a+3b)(3b+4c)(2c+a)}=
2(2a+3b)(3b+4c)(2c+a)
This is the required answer.
Answer:
This is picture of park and many peoples are playing and walking a park