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Hi ,
cubeth root of ( x + 1 ) ( y + 2 ) ( z + 3 ) = 7
do the cube of both sides of the equation.
( x + 1 ) ( y + 2 ) ( z + 3 ) = 7³----( 1 )
let us assume ,
( 6 + 1 ) ( 5 + 2 ) ( 4 + 3 ) = 7³ ---( 2 )
compare ( 1 ) and ( 2 ) , we get
x = 6 , y = 5 , z = 4
Now ,
x × y × z = 6 × 5 × 4
= 120
I hope this helps you.
: )
cubeth root of ( x + 1 ) ( y + 2 ) ( z + 3 ) = 7
do the cube of both sides of the equation.
( x + 1 ) ( y + 2 ) ( z + 3 ) = 7³----( 1 )
let us assume ,
( 6 + 1 ) ( 5 + 2 ) ( 4 + 3 ) = 7³ ---( 2 )
compare ( 1 ) and ( 2 ) , we get
x = 6 , y = 5 , z = 4
Now ,
x × y × z = 6 × 5 × 4
= 120
I hope this helps you.
: )
Answered by
0
Answer:
((x+1)*(y+2)*(z+3))⅓=7
=> x = 6 , y = 5, z = 4
=> ((6+1)*(5+2)*(4+3))⅓=7
=> (7*7*7)⅓=7
=> (7³)⅓=7
=> 7=7
=> xyz = 6*5*4 = 120
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