If 2a - 3b + 4c = 0 , then show that 8a³ - 27b³ + 64c³ + 72abc = 0
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Answers
Answer:
We have,
2a+3b+c=0
2a+3b=−c …….. (1)
On taking cube both sides, we get
(2a+3b)
3
=(−c)
3
8a
3
+27b
3
+3×2a×3b(2a+3b)=−c
3
8a
3
+27b
3
+18ab(2a+3b)=−c
3
From equation (1),
8a
3
+27b
3
+18ab(−c)=−c
3
8a
3
+27b
3
−18abc=−c
3
8a
3
+27b
3
+c
3
=18abc
Hence, proved.
Step-by-step explanation:
Given :-
2a -3b +4c = 0
To find:-
Show that 8a^3 -27b^3 +64c^3+72abc = 0
Solution:-
Given that
2a -3b +4c = 0 -------------(1)
Method-1:-
We know that
If a + b+ c = 0then a^3+b^3+c^3 = 3abc
On applying this to the equation (1)
Where a = 2a; b = -3b ; c = 4c
Now ,
If 2a -3b +4c = 0 then (2a)^3+(-3b)^3 + (4c)^3 =3(2a)(-3b)(4c)
=> 8a^3 +(-27b^3)+(64c^3)= - 72abc
=> 8a^3 -27b^3+64c^3= -72abc
=>8a^3 -27b^3+64c^3+72abc = 0
Method-2:-
Given that
2a -3b +4c = 0
2a -3b = -4c -------------(1)
On cubing both sides then
=> (2a-3b)^3 = (-4c)^3
We know that
(a-b)^3 = a^3-3ab(a-b)-b^3
Where a = 2a and b= 3b
=> (2a)^3-3(2a)(3b)(2a-3b)-(3b)^3 = -64c^3
=> 8a^3-18ab(2a-3b)-27b^3 = -64c^3
=> 8a^3-18ab(-4c)-27b^3 = -64c^3
(from (1))
=> 8a^3+72abc-27b^3 = -64c^3
=> 8a^3+72abc-27b^3+64c^3 = 0
=> 8a^3-27b^3+64c^3+72abc = 0
Hence, Proved.
Answer:-
If 2a-3b+4c=0 then 8a^3-27b^3+64c^3+72abc=0
Used formulae:-
- (a-b)^3 = a^3-3ab(a-b)-b^3
- If a + b+ c = 0then a^3+b^3+c^3 = 3abc