Suppose a⋄ba \diamond ba⋄b means ab+1ab + 1ab+1 and a∙ba \bullet ba∙b means a+ba + ba+b. Evaluate 4⋄((6∙8)∙(3⋄5)).4 \diamond ((6 \bullet 8) \bullet (3 \diamond 5)).4⋄((6∙8)∙(3⋄5)). Solution 4⋄((6∙8)∙(3⋄5))=4⋄((6+8)∙(3⋅5+1))=4⋄(14∙16)=4⋄(14+16)=4⋄30=4⋅30+1=121. \begin{aligned} 4 \diamond ((6 \bullet 8) \bullet (3 \diamond 5)) &=& 4 \diamond ((6 + 8) \bullet (3 \cdot 5 + 1))\\ &=& 4 \diamond (14 \bullet 16)\\ &=& 4 \diamond (14 + 16)\\ &=& 4 \diamond 30\\ &=& 4 \cdot 30 + 1\\ &=& 121. \end{aligned} 4⋄((6∙8)∙(3⋄5))======4⋄((6+8)∙(3⋅5+1))4⋄(14∙16)4⋄(14+16)4⋄304⋅30+1121.