If a:b=b:c then show that (a2+ab+ b2)(b2-bc+c2)=a2b2+b2c2+c2a2.
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Hope this helps.
Given a : b = b : c <=> ac = b²
So
( a² + ab + b² ) ( b² - bc + c² )
= ( a² + ab + ac ) ( ac - bc + c² )
= ac ( a + b + c ) ( a - b + c )
= ac ( (a+c)² - b² )
= ac ( a² + 2ac + c² - ac )
= b² ( a² + c² + ac )
= a²b² + b²c² + b²ac
= a²b² + b²c² + a²c²
Papa2887:
can i ask one more question~
Of course!
If a,b,c are in continued proportion , then show that, a2b2c2(1/a3+1/b3+1/c3)=a3+b3+c3.
thank you for helping me
Given ac=b2. Then a2b2c2=b2(ac)2=b2(b2)2=b6. So LHS = b6(1/a3+1/b3+1/c3)=(b2/a)3+b6/b3+(b2/c)3=c3+b3+a3
thank you very much..Please tell if u need any help.
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