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If the root of the equation 3x³+Px³+Qx-37=0 are each one more than the roots of the equation x³-Ax²+Bx-C, where A,B,C,P & Q are constants, then the value of A+B+C is equal to :

Actually I solved this question by let a root of 3x³+Px³+Qx-37=0 is t and according to question root of x³-Ax²+Bx-C would be t-1 And they both root would be satisfied their respective equation as ;

3t³+Pt³+Qt-37=0 (t-1)³-A(t-1)²+B(t-1)-c=0 => t³-t²(3+A)+t(3+2A)-(A+B+C)-1=0 Now I multiplied 3 both side of eqn And got 3t³-3t²(3+A)+3t(3+2A)-3(A+B+C)-3=0

Then I put 37= 3(A+B+C)+3 34\3=A+B+C = 11.33 But I don't know that it my right approach to this question

Please tell me can we do this question like this ?

Soni
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1 Answers1

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An alternate approach:

Let $\alpha, \beta$ and $\gamma$ be the 3 roots of $x^3-Ax^2+Bx-C=0$,

then we have $$x^3-Ax^2+Bx-C=(x-\alpha)(x-\beta)(x-\gamma)\tag{1}\label{eq1}$$

and

$$x^3+\frac{P}{3}x^2+\frac{Q}{3}x-\frac{37}{3}=(x-\alpha-1)(x-\beta-1)(x-\gamma-1)\tag{2}\label{eq2}$$

In equation $(1)$, if we put $x=-1$, we have

$$-1-A-B-C=(-1-\alpha)(-1-\beta)(-1-\gamma)$$

$$A+B+C=-1-(-1-\alpha)(-1-\beta)(-1-\gamma)$$

Job is done if we can find the value of

$$(-1-\alpha)(-1-\beta)(-1-\gamma)$$

from equation (2)

Li Kwok Keung
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