Question

If alpha,beta and gamma are the zeroes of x^3-12x^2+39x-28 them
find the value of alpha^3+beta^3+gamma^3​

Answer 1

Answer :-

408

Solution :-

x³ - 12x² + 39x - 28

Comparing with ax³ + bx² + cx + d we get,

• a = 1
• b = - 12
• c = 39
• d = - 28

α, β and γ are the zeroes of the polynomial

• Sum of zeroes = α + β + γ = - b/a = - ( - 12 )/1 = 12

• Sum of product of zeroes taken two at a time = αβ + βγ + γα = c/a = 39/1 = 39

• Product of zeroes = αβγ = - d/a = - ( - 28 )/1 = 28

Finding α² + β² + γ²

Using identity α² + β² + γ² = (α + β + γ)² - 2(αβ + βγ + γα)

==> α² + β² + γ² = ( 12 )² - 2( 39 )

==> α² + β² + γ² = 144 - 78 = 66

Finding α³ + β³ + γ²

Using identity α³ + β³ + γ³ = (α + β + γ) { α² + β² + γ² - (αβ + βγ + γα) } + 3αβγ

==> α³ + β³ + γ³ = ( 12 ){ 66 - ( 39 ) } + 3( 28 )

==> α³ + β³ + γ³ = ( 12 )( 27 ) + 84

==> α³ + β³ + γ³ = 324 + 84

==> α³ + β³ + γ³ = 408

Therefore the value of α³ + β³ + γ³ is 408.

Answer 2

QUESTION :-

If alpha,beta and gamma are the zeroes of x³-12x²+39x-28 them

find the value of α³+β³+y³.

SOLUTION :-

➠ x³-12x²+39x-28

➠ Comparing we get,

➠ a = 1 , b = -12 , c = 39 , d = -28

➠ let ,the three zeroes be α , β & y.

➠ sum of zeroes = α+β+y

➠ ( -b/a)= -(-12/1) = 12.

➠ sum of product of zeroes = αβ +βy + αy

➠ (c/a) = ( 39/1) = 39.

➠ product of zeroes = α*β*y

➠ -( -d/a) = -(-28/1) = 28.

We know that,

➠ α²+β²+y² = (α+β+y) ²-2(αβ + βy+αy)

Now,

➠ α²+β²+y² = (12)²-2(39)

➠α²+β²+y² = 144 - 78

➠α²+β²+y² =66.

We know that,

➠ α³+β³+y³ = (α+β+y) (α²+β²+y²)-(αβ + βy+αy)+3αβy

Now,

➠α³+β³+y³ = (12) [(66--(39)] +3(28)

➠α³+β³+y³ = (12)(27) +84

➠α³+β³+y³ = 324 + 84

➠α³+β³+y³ = 408

HENCE, THE VALUE OF α³+β³+y³ IS 408.