factorise a^3/b^3+b^3/c^3+c^3/a^3-3 plz help
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Hi ,
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We know the algebraic identity ,
x³+y³+z³-3xyz
=(x+y+z)( x² + y² + z² - xy - yz - zx )
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Here ,
a³/b³ + b³/c³ + c³/a³ - 3
= ( a/b )³ + ( b/c )³ + ( c/a )³-3(a/b)(b/c)(c/a)
=(a/b+b/c+c/a)[(a/b)²+(b/c)²+(c/a)²-a/c-b/a-c/b]
I hope this helps you.
: )
***********************************
We know the algebraic identity ,
x³+y³+z³-3xyz
=(x+y+z)( x² + y² + z² - xy - yz - zx )
*******************************
Here ,
a³/b³ + b³/c³ + c³/a³ - 3
= ( a/b )³ + ( b/c )³ + ( c/a )³-3(a/b)(b/c)(c/a)
=(a/b+b/c+c/a)[(a/b)²+(b/c)²+(c/a)²-a/c-b/a-c/b]
I hope this helps you.
: )
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